Frank Solutions for Chapter 21 Trigonometric Identities Class 10 ICSE Mathematics

Exercise 21.1


1. Prove the following identities:

(i) (1 – sin2 Î¸) sec2θ = 1

(ii) (1 – cos2 Î¸) sec2θ = tan2θ

(iii) tan A + cot A = sec A cosec A

(iv) sin θ(1 + tan θ) + cos θ (1 + cot θ) = sec θ + cosec θ

(v) (1 + cot θ – cosec θ) (1 + tan θ + sec θ) = 2

(vi) sin θ cot θ + sin θ cosec θ = 1 + cos θ

(vii) sec A (1 – sin A) (sec A + tan A) = 1

(viii) sec A (1 + sin A) (sec A – tan A) = 1

(ix) cosec θ (1 + cos θ) (cosec θ – cot θ) = 1

(x) (sec A – 1)/(sec A + 1) = (1 – cos A)/(1 + cos A)

(xi) (1 + sin A)/(1 – sin A) = (cosec A + 1)/(cosec A – 1)

(xii) cos A/(1 + sin A) = sec A – tan A

(xiii) (tan θ + sec θ – 1)/(tan θ – sec + 1) = (1 + sin θ)/cos θ

Answer

(i) (1 – sin2 Î¸) sec2θ = 1

From the question first we consider Left Hand Side (LHS),

= (1 – sin2 Î¸) sec2θ

We know that, (1 – sin2 Î¸) = cos2 Î¸

= cos2θ sec2θ

Also we know that, sec2θ = 1/cos2θ

= cos2 Î¸ × (1/cos2θ)

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(ii) (1 – cos2 Î¸) sec2θ = tan2θ

From the question first we consider Left Hand Side (LHS),

= (1 – cos2 Î¸) sec2θ

We know that, (1 – cos2 Î¸) = sin2 Î¸

= sin2θ sec2θ

Also we know that, sec2θ = 1/cos2θ

= sin2 Î¸ × (1/cos2θ)

= sin2θ/cos2θ

= tan2θ

Then, Right Hand Side (RHS) = tan2θ

Therefore, LHS = RHS

(iii) tan A + cot A = sec A cosec A

From the question first we consider Left Hand Side (LHS),

= tan A + cot A

We know that, tan A = sin A/cos A, cot A = cos A/ sin A

Then,

= (sin A/cos A) + (cos A/sin A)

= (sin2 A + cos2 A)/(sin A cos A)

Also we know that, sin2 A + cos2 A = 1

= 1/(sin A cos A)

= (1/sin A)(1/cos A)

= cosec A sec A

Then, Right Hand Side (RHS) = sec A cosec A

Therefore, LHS = RHS

(iv) sin θ(1 + tan θ) + cos θ (1 + cot θ) = sec θ + cosec θ

From the question first we consider Left Hand Side (LHS),

= sin θ(1 + tan θ) + cos θ (1 + cot θ)

We know that, tan θ = sin θ/cos θ, cot θ = cos θ/sin θ

= sin θ(1 + (sin θ/cos θ)) + cos θ (1 + (cos θ/sin θ))

= sin θ((cos θ + sin θ)/cos θ)) + cos θ ((sin θ + cos θ)/sin θ))

= cos θ + sin θ ((sin θ/cos θ) + (cos θ/sin θ))

= cos θ + sin θ ((1/sin θ)(1/cos θ))

= sec θ + cosec θ

Then, Right Hand Side (RHS) = sec θ + cosec θ

Therefore, LHS = RHS

(v) (1 + cot θ – cosec θ) (1 + tan θ + sec θ) = 2

From the question first we consider Left Hand Side (LHS),

= (1 + cot θ – cosec θ) (1 + tan θ + sec θ)

We know that,

cot θ = sin θ/cos θ, cosec θ = 1/cos θ, tan θ = cos θ/sin θ, sec θ = 1/sin θ

= (1 + (sin θ/cos θ) + (1/cos θ)) (1 + (cos θ/sin θ) – (1/sin θ))

Taking LCM we get,

= ((cos θ + sin θ + 1)/cos θ) ((sin θ + cos θ – 1)/sin θ)

= ((sin θ + cos θ)2 – 12)/(sin θ cos θ)

= (1 + 2 sin θ cos θ – 1)/(sin θ cos θ)

= (2 sin θ cos θ)/(sin θ cos θ)

By simplification we get,

= 2

Then, Right Hand Side (RHS) = 2

Therefore, LHS = RHS

(vi) sin θ cot θ + sin θ cosec θ = 1 + cos θ

From the question first we consider Left Hand Side (LHS),

= sin θ cot θ + sin θ cosec θ

We know that, cot θ = cos θ/sin θ, cosec θ = 1/sin θ

= sin θ (cos θ/sin θ) + sin θ (1/sin θ)

= cos θ + 1

Then, Right Hand Side (RHS) = 1 + cos θ

Therefore, LHS = RHS

(vii) sec A (1 – sin A) (sec A + tan A) = 1

From the question first we consider Left Hand Side (LHS),

= sec A (1 – sin A) (sec A + tan A)

We know that, sec A = 1/cos A, tan A = sin A/cos A

= (1/cos A) (1 – sin A) ((1/cos A) + (sin A/cos A))

= ((1 – sin A)/cos A) ((1 + sin A)/cos A)

By simplification we get,

= (1 – sin2A)/cos2 A

= cos2 A/cos2 A

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(viii) sec A (1 + sin A) (sec A – tan A) = 1

From the question first we consider Left Hand Side (LHS),

= sec A (1 + sin A) (sec A – tan A)

We know that, sec A = 1/cos A, tan A = sin A/cos A

= (1/cos A) (1 + sin A) ((1/cos A) – (sin A/cos A))

= ((1 + sin A)/cos A) ((1 – sin A)/cos A)

By simplification we get,

= (1 – sin2A)/cos2 A

= cos2 A/cos2 A

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(ix) cosec θ (1 + cos θ) (cosec θ – cot θ) = 1

From the question first we consider Left Hand Side (LHS),

= cosec θ (1 + cos θ) (cosec θ – cot θ)

We know that, cosec = 1/sin θ, cot θ = cos θ/sin θ

= (1/sin θ) (1 + cos θ) ((1/sin θ) – (cos θ/sin θ))

= ((1 + cos θ)/sin θ) ((1 – cos θ)/sin θ)

= ((1 – cos2 Î¸)/sin2 Î¸)

= sin2 Î¸/sin2 Î¸

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(x) (sec A – 1)/(sec A + 1) = (1 – cos A)/(1 + cos A)

From the question first we consider Left Hand Side (LHS),

= (sec A – 1)/(sec A + 1)

We know that, sec A = 1/cos A

= ((1/cos A) – 1)/((1/cos A) + 1)

By simplification we get,

= (1 – cos A)/(1 + cos A)

Then, Right Hand Side (RHS) = (1 – cos A)/(1 + cos A)

Therefore, LHS = RHS

(xi) (1 + sin A)/(1 – sin A) = (cosec A + 1)/(cosec A – 1)

From the question first we consider Left Hand Side (LHS),

= (1 + sin A)/(1 – sin A)

Then consider Right Hand Side (RHS) = (cosec A + 1)/(cosec A – 1)

We know that, cosec A = 1/sin A

So,

= ((1/sin A) + 1)/((1/sin A) – 1)

= (1 + sin A)/(1 – sin A)

Therefore, LHS = RHS

(xii) cos A/(1 + sin A) = sec A – tan A

From the question first we consider Right Hand Side (RHS),

= sec A – tan A

We know that, sec A = 1/cos A, tan A = sin A/cos A

= (1/cos A) – (sin A/cos A)

= (1 – sin A)/cos A

= ((1 – sin A)/cos A) ((1 + sin A)/(1 + sin A))

By cross multiplication we get,

= (1 – sin2 A)/(cos A(1 + sin A))

= cos2 A/(cos A (1 + sin A))

= cos A/(1 + sin A)

Then, Left Hand Side (LHS) = cos A/(1 + sin A)

Therefore, LHS = RHS

(xiii) (tan θ + sec θ – 1)/(tan θ – sec + 1) = (1 + sin θ)/cos θ

From the question first we consider Left Hand Side (LHS),

= (tan θ + sec θ – 1)/(tan θ – sec + 1)

The above terms can be written as,

= (tan θ + sec θ – (sec2 Î¸ – tan2 Î¸))/(1 + tan θ – sec θ)

= [tan θ + sec θ – {(sec θ + tan θ) (sec θ – tan θ)}]/(1 + tan θ – sec θ)

= [(tan θ + sec θ) {1 – (sec θ – tan θ)}]/(1 + tan θ – sec θ)

= [(tan θ + sec θ) (1 + tan θ – sec θ)]/(1 + tan θ – sec θ)

= [tan θ + sec θ]

= (1 + sin θ)/ cos θ

Then, Right Hand Side (RHS) = (1 + sin θ)/cos θ

Therefore, LHS = RHS


2. Prove the following identities:

(i) sin2 A cos2 B – cos2 A sin2 B = sin2 A – sin2 B

(ii). (1 – tan A)2 + (1 + tan A)2 = 2 sec2 A

(iii) cosec4 A – cosec2 A = cot4 A + cot4 A

(iv) sec2 A + cosec2 A = sec2 A cosec2 A

(v) cos4 A – sin4 A = 2 cos2 A – 1

(vi) (sec A – cos A) (sec A + cos A) = sin2 A + tan2 A

(vii) (cos A + sin A)2 + (cos A – sin A)2 = 2

(viii) (cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1

(ix) sec2 A cosec2 A = tan2 A + cot2 A + 2

(x) (cos A/(1 – tan A)) + (sin A/(1 – cot A)) = sin A + cos A

(xi) (cosec A – sin A) (sec A – cos A) = 1/(tan A + cot A)

(xii) sin4 A + cos4 A = 1 – 2 sin2 A cos2 A

Answer

(i) sin2 A cos2 B – cos2 A sin2 B = sin2 A – sin2 B

From the question first we consider Left Hand Side (LHS),

= sin2 A cos2 B – cos2 A sin2 B

We know that, cos2 B = (1 – sin2 B),

= sin2 A (1 – sin2 B) – (1 – sin2 A) sin2 B

= sin2 A – sin2 A sin2 B – sin2 B + sin2 A sin2 B

On simplification we get,

= sin2 A – sin2 B

Then, Right Hand Side (RHS) = sin2 A – sin2 B

Therefore, LHS = RHS

(ii) (1 – tan A)2 + (1 + tan A)2 = 2 sec2 A

From the question first we consider Left Hand Side (LHS),

= (1 – tan A)2 + (1 + tan A)2

Expanding the above terms we get,

= 1 + tan2 A – 2 tan A + 1 + tan2 A + 2 tan A

= 2(1 + tan2 A)

= 2 sec2 A

Then, Right Hand Side (RHS) = 2 sec2 A

Therefore, LHS = RHS

(iii) cosec4 A – cosec2 A = cot4 A + cot4 A

From the question first we consider Left Hand Side (LHS),

= cosec4 A – cosec2 A

By taking common we get,

= cosec2 A (cosec2 A – 1)

Now we consider Right Hand Side (RHS) = cot4 A + cot4 A

Again taking common we get,

= cot2 A (cot2 A + 1)

We know that, cot2 A = cosec2 A – 1, cot2 A + 1 = cosec2 A

So, (cosec2 A – 1) cosec2 A

Therefore, LHS = RHS

(iv) sec2 A + cosec2 A = sec2 A cosec2 A

From the question first we consider Left Hand Side (LHS),

= sec2 A + cosec2 A

We know that, sec2 A = 1/cos2 A, cosec2 A = 1/sin2 A

= (1/cos2 A) + (1/sin2 A)

= (sin2 A + cos2 A)/(cos2 A sin2 A)

= 1/(cos2 A sin2)

= sec2 A cosec2 A

Then, Right Hand Side (RHS) = sec2 A cosec2 A

Therefore, LHS = RHS

(v) cos4 A – sin4 A = 2 cos2 A – 1

From the question first we consider Left Hand Side (LHS),

= cos4 A – sin4 A

We know that, a2 – b2 = (a + b) (a – b)

= (cos2 A – sin2 A)(cos2 A + sin2 A)

= (cos2 A – (1 – cos2 A))

= 2 cos2 A – 1

Then, Right Hand Side (RHS) = 2 cos2 A – 1

Therefore, LHS = RHS

(vi) (sec A – cos A) (sec A + cos A) = sin2 A + tan2 A

From the question first we consider Left Hand Side (LHS),

= (sec A – cos A) (sec A + cos A)

We know that, a2 – b2 = (a + b) (a – b)

= (sec2 A – cos2 A)

Also we know that, sec2 A = 1 + tan2 A, cos2 A = 1 – sin2 A

= 1 + tan2 A – (1 – sin2 A)

= tan2 A + sin2 A

Then, Right Hand Side (RHS) = tan2 A + sin2 A

Therefore, LHS = RHS

(vii) (cos A + sin A)2 + (cos A – sin A)2 = 2

From the question first we consider Left Hand Side (LHS),

= (cos A + sin A)2 + (cos A – sin A)2

We know that, (a + b)2 = a2 + 2ab + b2, (a – b)2 = a2 – 2ab + b2

= cos2 A + sin2 A + 2 cos A sin A + cos2 A + sin2 A – 2 cos A sin A

By simplification we get,

= 2(cos2 A + sin2 A)

= 2

Then, Right Hand Side (RHS) = 2

Therefore, LHS = RHS

(viii) (cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1

From the question first we consider Left Hand Side (LHS),

= (cosec A – sin A) (sec A – cos A) (tan A + cot A)

We know that, cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A

= ((1/sin A) – sin A) ((1/cos A) – cos A) (tan A + (1/tan A))

= ((1 – sin2 A)/sin A) ((1 – cos2 A)/cos A) ((sin A/cos A) + (cos A/sin A))

= (cos2 A/sin A) (sin2 A/cos A) ((sin2 A + cos2 A)/(sin A cos A))

By simplification we get,

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(ix) sec2 A cosec2 A = tan2 A + cot2 A + 2

From the question first we consider Left Hand Side (LHS),

= sec2 A cosec2 A

We know that, sec2 A = 1/cos2 A, cosec2 A = 1/sin2 A

= 1/(cos2 A sin2 A)

Now consider RHS = tan2 A + cot2 A + 2

= tan2 A + cot2 A + 2 tan2 A cot2 A

= (tan A + cot A)2

= ((sin A/cos A) + (cos A/sin A))2

= ((sin2 A + cos2 A)/(sin A cos A))

= 1/(cos2 A sin2 A)

Therefore, LHS = RHS

(x) (cos A/(1 – tan A)) + (sin A/(1 – cot A)) = sin A + cos A

From the question first we consider Left Hand Side (LHS),

= (cos A/(1 – tan A)) + (sin A/(1 – cot A))

= (cos A/(1 – (sin A/cos A))) + (sin A/(1 – (cos A/sin A)))

= (cos A/((cos A – sin A)/cos A)) + (sin A/((sin A – cos A)/sin A))

= (cos2A/(cos A – sin A)) + (sin2A/(sin A – cos A))

= (cos2 A – sin2 A)/(cos A – sin A)

= sin A + cos A

Then, Right Hand Side (RHS) = sin A + cos A

Therefore, LHS = RHS

(xi) (cosec A – sin A) (sec A – cos A) = 1/(tan A + cot A)

From the question first we consider Left Hand Side (LHS),

= (cosec A – sin A) (sec A – cos A)

We know that, cosec A = 1/sin A, sec A = 1/cos A

= ((1/sin A) – sin A) ((1/cos) – cos A)

= ((1 – sin2 A)/sin A)((1 – cos2 A)/cos A)

= (cos2 A/sin A) (sin2 A/cos A)

= cos A sin A

Now consider the RHS = 1/(tan A + cot A)

= (1/((sin A/cos A)+(cos A/sin A))

= 1/((sin2 A + cos2 A)/(sin A cos A))

= cos A sin A

Therefore, LHS = RHS

(xii) sin4 A + cos4 A = 1 – 2 sin2 A cos2 A

From the question first we consider Left Hand Side (LHS),

= sin4 A + cos4 A

= 1 – 2 sin2 A cos2 A

= sin4 A + cos4 A + 2 sin2 A cos2 A

= (sin2A)2+(cos2A)2+ 2 sin2A cos2A − 2sin2Acos2A

[Adding and subtracting 2 sin2A cos2A]

= (sin2A + cos2A)2 – 2 sin2 A cos2 A

= 1 – 2 sin2 A cos2 A

Then, Right Hand Side (RHS) = 1 – 2 sin2 A cos2 A

Therefore, LHS = RHS


3. Prove the following identities:

(i) (sin A/(1 + cos A)) + ((1 + cos A)/sin A) = 2 cosec A

(ii) (1 + cos A)/(1 – cos A) = (cosec A + cot A)2

(iii) (cot A + tan B)/(cot B + tan A) = cot A tan B

(iv) 1/(tan A + cot A) = sin A cos A

(v) tan A – cot A = (1 – 2 cos2 A)/(sin A cos A)

(vi) ((1 + tan2 A) cot A)/cosec2 A = tan A

(vii) cosec A + cot A = (1/(cosec A – cot A))

(viii) ((cosec A)/(cosec A – 1)) + ((cosec A)/(cosec A + 1)) = 2 sec2 A

(ix) (1 + cos A)/(1 – cos A) = (tan2 A)/(sec A – 1)2

(x) (cot A – cosec A)2 = ((1 – cos A)/(1 + cos A))

Answer

(i) (sin A/(1 + cos A)) + ((1 + cos A)/sin A) = 2 cosec A

From the question first we consider Left Hand Side (LHS),

= (sin A/(1 + cos A)) + ((1 + cos A)/sin A)

By cross multiplication we get,

= ((1 + cos A)2 + sin2 A)/(sin A (1 + cos))

We know that, (a + b)2 = a2 + 2ab + b2

Then,

= (1 + 2 cos A + cos2 A + sin2 A)/(sin A (1 + cos A)

= (2 + 2 cos A)/(sin A (1 + cos A))

Now taking common outside we get,

= (2(1 + cos A))/(sin A (1 + cos A))

= 2 cosec A

Then, Right Hand Side (RHS) = 2 cosec A

Therefore, LHS = RHS

(ii) (1 + cos A)/(1 – cos A) = (cosec A + cot A)2

From the question first we consider Left Hand Side (LHS),

= (1 + cos A)/(1 – cos A)

Now, multiply and divide by (1 + cos A) we get,

= ((1 + cos A)/(1 – cos A)) ((1 + cos A)/(1 + cos A))

By cross multiplication we get,

= (1 + cos A)2/(1 – cos2 A)

We know that, sin2 A + cos2 A = 1

So, (1 + cos A)2/sin2 A

= ((1 + cos A)/sin A)2

= ((1/sin A) + (cos A/sin A))2

= (cosec A + cot A)2

Then, Right Hand Side (RHS) = (cosec A + cot A)2

Therefore, LHS = RHS

(iii) (cot A + tan B)/(cot B + tan A) = cot A tan B

From the question first we consider Left Hand Side (LHS),

= (cot A + tan B)/(cot B + tan A)

We know that, cot A = 1/tan A

= ((1/tan A) + (tan B))/((1/tan A) + tan A)

= ((1 + tan A tan B)/tan A)/((1 + tan A tan B)/tan B)

= ((1 + tan A tan B)/tan A) (tan B/(1 + tan A tan B))

= tan B/tan A

= (1/tan A) tan B

= cot A tan B

Then, Right Hand Side (RHS) = cot A tan B

Therefore, LHS = RHS

(iv) 1/(tan A + cot A) = sin A cos A

From the question first we consider Left Hand Side (LHS),

= 1/(tan A + cot A)

We know that, tan A = sin A/cos A, cot A = cos A/sin A

= 1/((sin A/cos A) + (cos A/sin A))

By cross multiplication we get,

= 1/((sin2 A + cos2 A)/(sin A cos A))

Also we know that, sin2 A + cos2 A = 1

= 1/(1/(sin A cos A))

= sin A cos A

Then, Right Hand Side (RHS) = sin A cos A

Therefore, LHS = RHS

(v) tan A – cot A = (1 – 2 cos2 A)/(sin A cos A)

From the question first we consider Left Hand Side (LHS),

= tan A – cot A

We know that, tan A = sin A/cos A, cot A = cos A/sin A

Then,

= (sin A/cos A) – (cos A /sin A)

= (sin2 A – cos2 A)/(sin A cos A)

We know that, sin2 A = 1 – cos2 A

= (1 – cos2 A – cos2 A)/(sin A cos A)

So,

= (1 – 2 cos2 A)/(sin A cos A)

Then, Right Hand Side (RHS) = (1 – 2 cos2 A)/(sin A cos A)

Therefore, LHS = RHS

(vi) ((1 + tan2 A) cot A)/cosec2 A = tan A

From the question first we consider Left Hand Side (LHS),

= ((1 + tan2 A) cot A)/cosec2 A

We know that, 1 + tan2 A = sec2

= (sec2 cot A)/cosec2 A

Also we know that, sec2 A = 1/cos2 A, cot A = cos A/sin A

= ((1/cos2 A)(cos A/sin A))/(1/(sin2 A))

= (1/(cos A sin A))/(1/sin2 A)

= sin A/cos A

= tan A

Then, Right Hand Side (RHS) = tan A

Therefore, LHS = RHS

(vii) cosec A + cot A = (1/(cosec A – cot A))

From the question first we consider Left Hand Side (LHS),

= cosec A + cot A

Now multiply and divide by cosec A – cot A we get,

= ((cosec A + cot A)/1) ((cosec A – cot A)/(cosec A – cot A))

By cross multiplication we get,

= (cosec2 A – cot2 A)/(cosec A – cot A)

= (1 + cot2 A – cot2 A)/(cosec A – cot A)

= (1/(cosec A – cot A))

Then, Right Hand Side (RHS) = (1/(cosec A – cot A))

Therefore, LHS = RHS

(viii) ((cosec A)/(cosec A – 1)) + ((cosec A)/(cosec A + 1)) = 2 sec2 A

From the question first we consider Left Hand Side (LHS),

= ((cosec A)/(cosec A – 1)) + ((cosec A)/(cosec A + 1))

By cross multiplication we get,

= (cosec2 A + cosec A + cosec2 A – cosec A)/(cosec2 A – 1)

We know that, cosec2 A – 1 = cot2 A

= 2 cosec2 A/cot2 A

Also we know that, cosec2 A = 1/sin2 A

= (2/sin2 A)/(cos2 A/sin2 A)

= 2/cos2 A

= 2 sec2 A

Then, Right Hand Side (RHS) = 2 sec2 A

Therefore, LHS = RHS

(ix) (1 + cos A)/(1 – cos A) = (tan2 A)/(sec A – 1)2

From the question first we consider Left Hand Side (LHS),

= (1 + cos A)/(1 – cos A)

We know that, cos A = 1/sec A

Then,

= (1 + (1/sec A))/(1 – (1/sec))

= (sec A + 1)/(sec A – 1)

Now multiply and divide by (sec A – 1) we get,

= ((sec A + 1)/(sec A – 1)) ((sec A – 1)/(sec A – 1))

By cross multiplication we get,

= (sec2 A – 1)/(sec A – 1)2

= tan2 A/(sec A – 1)2

Then, Right Hand Side (RHS) = tan2 A/(sec A – 1)2

Therefore, LHS = RHS

(x) (cot A – cosec A)2 = ((1 – cos A)/(1 + cos A))

From the question first we consider Left Hand Side (LHS),

= (cot A – cosec A)2

We know that, cot A = cos A/sin A, cosec A = 1/sin A

= ((cos A/sin A) – (1/sin A))2

= ((cos A – 1)/sin A)2

= (cos A – 1)2/sin2 A

= (cos A – 1)2/1 – cos2 A

= (-(1 – cos A)2/((1 – cos A) (1 + cos A))

= ((1 – cos A) (1 – cos A))/((1 – cos A)(1 + cos A))

= (1 – cos A)/(1 + cos A)

Then, Right Hand Side (RHS) = (1 – cos A)/(1 + cos A)

Therefore, LHS = RHS


4. Prove the following identities:

(i) √(cosec2 q – 1) = cos q cosec q

(ii) √((1 + sin q)/(1 – sin q)) + √((1 – sin q)/(1 + sin q)) = 2 sec q

(iii) √((1 – cos A)/(1 + cos A)) = (sin A/(1 + cos A))

(iv) √((1 + cos A)/(1 – cos A)) = cosec A + cot A

(v) √((sec q – 1)/(sec q + 1)) + √((sec q + 1)/(sec q – 1)) = 2 cosec q

Answer






5. Prove the following identities:

(i) (sec θ – tan θ)2 = (1 – sin θ)/(1 + sin θ)

(ii) (1/(sin A + cos A)) + (1/(sin A – cos A)) = 2 sin A/(1 – 2 cos2 A)

(iii) ((sin A + cos A)/(sin A – cos A)) + ((sin A – cos A)/(sin A + cos A)) = 2 /(2 sin2 A – 1)

(iv) tan2 A – tan2 B = (sin2A – sin2 B)/(cos2 A cos2 B)

(v) (cos A/(1 – tan A)) + (sin2 A/(sin A – cos A)) = cos A + sin A

(vi) (1 + tan2 A) + (1 + (1/tan2 A)) = (1/(sin2 A – sin4 A))

(vii) ((cos3 A + sin3 A)/(cos A + sin A)) + ((cos3 A – sin3 A)/(cos A – sin A)) = 2

(viii) (tan θ + (1/cos θ))2 + (tan θ – (1/cos θ))2 = 2((1 + sin2 Î¸)/(1 – sin2 Î¸)

(ix) ((sin A – sin B)/(cos A + cos B)) + ((cos A – cos B)/(sin A + sin B)) = 0

(x) (1/(cos A + sin A – 1)) + (1/(cos A + sin A + 1)) = cosec A + sec A

(xi) (cot A + cosec A – 1)/(cot A – cosec A + 1) = (cos A + 1)/sin A

(xii) (sec A – 1)/(sec A + 1) = (sin2 A)/(1 + cos A)2

Answer

(i) (sec θ – tan θ)2 = (1 – sin θ)/(1 + sin θ)

From the question first we consider Left Hand Side (LHS),

= (sec θ – tan θ)2

We know that, sec θ = 1/cos θ, tan θ = sin θ/cos θ

= ((1/cos θ) – (sin θ/cos θ))2

By taking LCM we get,

= ((1 – sin θ)/cos θ)2

= (1 – sin θ)2/cos2 Î¸

Also we know that, cos2 Î¸ = 1 – sin2 Î¸

= (1 – sin θ)2/(1 – sin2 Î¸)

= (1 – sin θ)2/((1 – sin θ)(1 + sin θ))

= (1 – sin θ)/(1 + sin θ)

Then, Right Hand Side (1 – sin θ)/(1 + sin θ)

Therefore, LHS = RHS

(ii) (1/(sin A + cos A)) + (1/(sin A – cos A)) = 2 sin A/(1 – 2 cos2 A)

From the question first we consider Left Hand Side (LHS),

= (1/(sin A + cos A)) + (1/(sin A – cos A))

By taking LCM we get,

= (sin A – cos A + sin A + cos A)/(sin2 A – cos2 A)

= 2 sin A/(1 – cos2 A – cos2 A)

= 2 sin A/(1 – 2 cos2 A)

Then, Right Hand Side 2 sin A/(1 – 2 cos2 A)

Therefore, LHS = RHS

(iii) ((sin A + cos A)/(sin A – cos A)) + ((sin A – cos A)/(sin A + cos A)) = 2 /(2 sin2 A – 1)

From the question first we consider Left Hand Side (LHS),

= ((sin A + cos A)/(sin A – cos A)) + ((sin A – cos A)/(sin A + cos A))

By taking LCM we get,

= ((sin A + cos A)2 + (sin A – cos A)2)/((sin A + cos A)(sin A – cos A))

Then,

= (sin2 A + cos2 A + 2 sin A cos A + sin2 A + cos2 A – 2 sin cos A)/(sin2 A – cos2 A)

= (2 sin2 A + 2 cos2 A + 2 sin A cos A – 2 sin cos A)/(sin2 A – cos2 A)

= (2 sin2 A + 2 cos2 A)/(sin2 A – cos2 A)

= 2(sin2 A + cos2 A)/(sin2 A – cos2 A)

We know that, sin2 A + cos2 A = 1

= 2(1)/(sin2 A – cos2 A)

= 2/(sin2 A – cos2 A)

= 2/(sin2 A – (1 – sin2 A))

= 2/(2 sin2 A – 1)

Then, Right Hand Side 2/(2 sin2 A – 1)

Therefore, LHS = RHS

(iv) tan2 A – tan2 B = (sin2A – sin2 B)/(cos2 A cos2 B)

From the question first we consider Left Hand Side (LHS),

= tan2 A – tan2 B

We know that, tan2 Î¸ = sin2 Î¸ /cos2 Î¸,

= (sin2 A cos2 B – cos2 A sin2 B)/(cos2 A cos2 B)

Then,

= (((1 – cos2 A) cos2 B) – (cos2 A sin2 B))/(cos2 A cos2 B)

= (cos2 B – cos2 A cos2 B – cos2 A + cos2 A cos2 B)/(cos2 A cos2 B)

By simplification we get,

= (cos2 B – cos2 A)/(cos2 A cos2 B)

= ((1 – sin2 B) – (1 – sin2 A))/(cos2 A cos2 B)

= (sin2 A – sin2 B)/(cos2 A cos2 B)

Then, Right Hand Side (sin2 A – sin2 B)/(cos2 A cos2 B)

Therefore, LHS = RHS

(v) (cos A/(1 – tan A)) + (sin2 A/(sin A – cos A)) = cos A + sin A

From the question first we consider Left Hand Side (LHS),

= (cos A/(1 – tan A)) + (sin2 A/(sin A – cos A))

We know that, tan A = sin A/cos A

= (cos A/(1 – (sin A/cos A))) + (sin2 A/(sin A – cos A))

Then,

= (cos A/((cos A – sin A)/cos A)) + (sin2 A/(sin A – cos A))

= (cos2 A/(cos A – sin A)) – (sin2 A/(cos A – sin A))

= (cos2 A – sin2 A)/(cos A – sin A)

= ((cos A + sin A)(cos A – sin A))/(cos A – sin A)

By simplification we get,

= cos A + sin A

Then, Right Hand Side cos A + sin A

Therefore, LHS = RHS

(vi) (1 + tan2 A) + (1 + (1/tan2 A)) = (1/(sin2 A – sin4 A))

From the question first we consider Left Hand Side (LHS),

= (1 + tan2 A) + (1 + (1/tan2 A))

We know that, tan2 A = sin2 A/cos2 A

= (1 + (sin2 A/cos2 A)) + (1 + (1/(sin2A/cos2 A)))

By taking LCM we get,

= ((cos2 A + sin2 A)/cos2 A) + ((cos2 A + sin2 A)/sin2 A)

Also we know that, cos2 A + sin2 A = 1

So,

= (1/(1 – sin2 A)) + (1/sin2 A)

= (sin2 A + 1 – sin2 A)/(sin2 A(1 – sin2 A))

= 1/(sin2 A – sin4 A)

Then, Right Hand Side = 1/(sin2 A – sin4 A)

Therefore, LHS = RHS

(vii) ((cos3 A + sin3 A)/(cos A + sin A)) + ((cos3 A – sin3 A)/(cos A – sin A)) = 2

From the question first we consider Left Hand Side (LHS),

= ((cos3 A + sin3 A)/(cos A + sin A)) + ((cos3 A – sin3 A)/(cos A – sin A))

By taking LCM we get,

= ((cos3 A + sin3 A)(cos A – sin A) + (cos3 A – sin3 A)(cos A + sin A))/(cos2 A – sin2 A)

= 2(cos4 A – sin4 A)/(cos2 A – sin2 A)

= (2(cos2 A + sin2 A)(cos2 A – sin2 A))/(cos2 A – sin2 A)

= 2(cos2 A + sin2 A)

We know that, cos2 A + sin2 A = 1

= 2

Then, Right Hand Side = 2

Therefore, LHS = RHS

(viii) (tan θ + (1/cos θ))2 + (tan θ – (1/cos θ))2 = 2((1 + sin2 Î¸)/(1 – sin2 Î¸)

From the question first we consider Left Hand Side (LHS),

= (tan θ + (1/cos θ))2 + (tan θ – (1/cos θ))2

We know that, tan θ = sin θ/cos θ

= ((sin θ/cos θ) + (1/cos θ))2 + ((sin θ/cos θ) – (1/cos θ))2

= ((sin θ + 1)/cos θ)2 + ((sin θ – 1)/cos θ)2

= ((sin θ + 1)2/cos2 Î¸) + ((sin θ – 1)/cos2 Î¸)

= ((sin θ + 1)2 + (sin θ – 1)2)/cos2 Î¸

Also we know that, (a + b)2 = a2 + 2ab + b2

= (sin2 Î¸ + 1 + 2 sin θ + sin2 Î¸ + 1 – 2 sin θ)/(1 – sin2 Î¸)

= 2(1 + sin2 Î¸)/(1 – sin2 Î¸)

Then, Right Hand Side = 2(1 + sin2 Î¸)/(1 – sin2 Î¸)

Therefore, LHS = RHS

(ix) ((sin A – sin B)/(cos A + cos B)) + ((cos A – cos B)/(sin A + sin B)) = 0

From the question first we consider Left Hand Side (LHS),

= ((sin A – sin B)/(cos A + cos B)) + ((cos A – cos B)/(sin A + sin B))

By taking LCM we get,

= (((sin A + sin B)(sin A – sin B)) + ((cos A + cos B)(cos A – cos B)))/((cos A + cos B)(sin A – sin B))

By simplification we get,

= ((sin2 A – sin2 B) + (cos2 A – cos2 B))/((cos A + cos B)(sin A – sin B))

= ((sin2 A + cos2 A) – (sin2 B + cos2 B))/((cos A + cos B)(sin A – sin B))

We know that, sin2 A + cos2 A = 1

= (1 – 1)/ ((cos A + cos B)(sin A – sin B))

= 0/((cos A + cos B)(sin A – sin B))

= 0

Then, Right Hand Side = 0

Therefore, LHS = RHS

(x) (1/(cos A + sin A – 1)) + (1/(cos A + sin A + 1)) = cosec A + sec A

From the question first we consider Left Hand Side (LHS),

= (1/(cos A + sin A – 1)) + (1/(cos A + sin A + 1))

By taking LCM we get,

= (cos A + sin A + 1 + cos A + sin A – 1)/((cos A + sin A)2 – 1)

We know that, (a + b)2 = a2 + 2ab + b2

= (2(cos A + sin A))/(cos2 A + sin2 A + 2 cos A sin A – 1)

= (cos A + sin A)/(cos A sin A)

= (cos A/(cos A sin A)) + (sin A/(cos A sin A))

= (1/sin A) + (1/cos A)

= cosec A + sec A

Then, Right Hand Side = cosec A + sec A

Therefore, LHS = RHS

(xi) (cot A + cosec A – 1)/(cot A – cosec A + 1) = (cos A + 1)/sin A

From the question first we consider Left Hand Side (LHS),

= (cot A + cosec A – 1)/(cot A – cosec A + 1)

We know that, cosec2 A – cot2 A = 1

= (cot A + cosec – (cosec2 A – cot2 A))/(cot A – cosec A + 1)

Also we know that, (a2 – b2) = (a + b) (a – b)

= [cot A + cosec A – ((cosec A – cot A)(cosec A + cot A))]/(cot A – cosec A + 1)

= (cot A + cosec A [1 – cosec A + cot A])/(cot A – cosec A + 1)

= cot A + cosec A

= (cos A/sin A) + (1/sin A)

= (1 + cos A)/sin A

Then, Right Hand Side = (1 + cos A)/sin A

Therefore, LHS = RHS

(xii) (sec A – 1)/(sec A + 1) = (sin2 A)/(1 + cos A)2

From the question first we consider Left Hand Side (LHS),

= (sec A – 1)/(sec A + 1)

We know that, sec A = 1/cos A

= ((1/cos A) – 1)/((1/cos) + 1)

= (1 – cos A)/(1 + cos A)

Then,

= ((1 – cos A)/(1 + cos A)) × ((1 + cos A)/(1 + cos A))

By simplification we get,

= (1 – cos2 A)/(1 + cos A)2

Also we know that, 1 – cos2 A = sin2 A

= sin2 A/(1 + cos A)2

Then, Right Hand Side = sin2 A/(1 + cos A)2

Therefore, LHS = RHS


6. Prove the following identities:

(i) (1 + cot A)2 + (1 – cot A)2 = 2 cosec2 A

(ii) (cosec θ/(tan θ + cot θ)) = cos θ

(iii) (1 + tan2 Î¸) sin θ cos θ = tan θ

(iv) ((1 + sin θ)/(cosec θ – cot θ)) – ((1 – sin θ)/(cosec θ + cot θ)) = 2 (1 + cot θ)

(v) (1 + cot A + tan A) (sin A – cos A) = (sec A/cosec2 A) – (cosec A/sec2 A)

(vi) 2(sin6 Î¸ + cos6 Î¸) – 3 (sin4 Î¸ + cos4 Î¸) + 1 = 0

(vii) sin8 Î¸ – cos8 Î¸ = (sin2 Î¸ – cos2 Î¸)(1 – 2 sin2 Î¸ cos2 Î¸)

(viii) sec4 A – sec2 A = (sin2 A/cos4 A)

(ix) (tan2 Î¸/(tan2 Î¸ – 1)) + (cosec2 Î¸/(sec2 Î¸ – cosec2 Î¸)) = (1/(sin2 Î¸ – cos2 Î¸))

(x) (sec2 Î¸ – sin2 Î¸)/(tan2 Î¸) = cosec2 Î¸ – cos2 Î¸

Answer

(i) (1 + cot A)2 + (1 – cot A)2 = 2 cosec2 A

From the question first we consider Left Hand Side (LHS),

= (1 + cot A)2 + (1 – cot A)2

We know that, (a + b)2 = a2 + 2ab + b2

= 1 + cot2 A + 2 cot A + 1 + cot2 A – 2 cot A

= 2 + 2 cot2 A

Taking common terms outside we get,

= 2 (1 + cot2 A)

Also we know that, 1 + cot2 A = cosec2 A

= 2 cosec2 A

Then, Right Hand Side = 2 cosec2 A

Therefore, LHS = RHS

(ii) (cosec θ/(tan θ + cot θ)) = cos θ

From the question first we consider Left Hand Side (LHS),

= (cosec θ/(tan θ + cot θ))

We know that, cosec θ = 1/sin θ, tan θ = sin θ/cos θ, cot θ = cos θ/sin θ

Then,

= (1/sin θ)/((sin θ/cos θ) + (cos θ/sin θ))

Taking LCM in the denominator we get,

= (1/sin θ)/((sin2 Î¸ + cos2 Î¸)/(cos θ sin θ))

Also we know that, sin2 Î¸ + cos2 Î¸ = 1

= (1/sin θ)/(1/cos θ sin θ)

= (1/sin θ) × ((cos θ sin θ)/1)

= cos θ

Then, Right Hand Side = cos θ

Therefore, LHS = RHS

(iii) (1 + tan2 Î¸) sin θ cos θ = tan θ

From the question first we consider Left Hand Side (LHS),

= (1 + tan2 Î¸) sin θ cos θ

We know that, tan2 Î¸ = sin2 Î¸/cos2 Î¸

= (1 + (sin2 Î¸/cos2 Î¸)) sin θ cos θ

Taking LCM we get,

= ((cos2 Î¸ + sin2 Î¸)/cos2 Î¸) sin θ cos θ

Also we know that, sin2 Î¸ + cos2 Î¸ = 1

= (1/cos2 Î¸) sin θ cos θ

= sin θ/cos θ

= tan θ

Then, Right Hand Side = tan θ

Therefore, LHS = RHS

(iv) ((1 + sin θ)/(cosec θ – cot θ)) – ((1 – sin θ)/(cosec θ + cot θ)) = 2 (1 + cot θ)

From the question first we consider Left Hand Side (LHS),

= ((1 + sin θ)/(cosec θ – cot θ)) – ((1 – sin θ)/(cosec θ + cot θ))

By taking LCM we get,

= [((1 + sin θ)(cosec θ + cot θ)) – ((1 – sin θ)(cosec θ – cot θ))]/(cosec2 Î¸ – cot2 Î¸)

We know that, 1 + cot2 Î¸ = cosec2 Î¸

= (cosec θ + cot θ + 1 + cos θ – cosec θ + cot θ + 1 – cos θ)/(1 + cot2 Î¸ – cot2 Î¸)

By simplification we get,

= 2 + 2 cot θ

Taking common terms outside we get,

= 2(1 + cot θ)

Then, Right Hand Side = 2(1 + cot θ)

Therefore, LHS = RHS

(v) (1 + cot A + tan A) (sin A – cos A) = (sec A/cosec2 A) – (cosec A/sec2 A)

From the question first we consider Left Hand Side (LHS),

= (1 + cot A + tan A) (sin A – cos A)

We know that, cot A = cos A/sin A, tan A = sin A/cos A

= [1 + (cos A/sin A) + (sin A/cos A)] (sin A – cos A)

By taking LCM we get,

= [(sin A cos A + cos2 A + sin2 A)/(sin A cos A)] (sin A – cos A)

Also we know that, (a3 – b3) = (a – b) (a2 + b2 + ab)

So,

= (sin3 A – cos3 A)/(sin A cos A)

= (sin3 A/(sin A cos A)) – (cos3 A/(sin A cos A))

= (sin2 A/cos A) – (cos2 A/sin A)

= ((1/cos A) × sin2 A) – ((1/sin A) × cos2 A)

= sec A sin2 A – cosec A cos2 A

= (sec A/cosec2 A) – (cosec A/sec2 A)

Then, Right Hand Side = (sec A/cosec2 A) – (cosec A/sec2 A)

Therefore, LHS = RHS

(vi) 2(sin6 Î¸ + cos6 Î¸) – 3 (sin4 Î¸ + cos4 Î¸) + 1 = 0

From the question first we consider Left Hand Side (LHS),

= 2(sin6 Î¸ + cos6 Î¸) – 3 (sin4 Î¸ + cos4 Î¸) + 1

The above terms can be written as,

= 2[(sin2 Î¸)3 + (cos2 Î¸)3] – 3 (sin4 Î¸ + cos4 Î¸) + 1

We know that, (a3 + b3) = (a + b) (a2 + b2 – ab)

= 2[(sin2 Î¸ + cos2 Î¸) ((sin2 Î¸) + (cos2 Î¸)2 – sin2 Î¸ cos2 Î¸)] – 3(sin4 Î¸ + cos4 Î¸) + 1

= 2[(sin2 Î¸)2 + (cos2 Î¸)2 – sin2 Î¸ cos2 Î¸] – 3 (sin4 Î¸ + cos4 Î¸) + 1

= 2 sin4 Î¸ + 2 cos4 Î¸ – 2 sin2 Î¸ cos2 Î¸ – 3 sin4 Î¸ – 3 cos4 Î¸ + 1

= – sin4 Î¸ – cos4 Î¸ – 2 sin2 Î¸ cos2 Î¸ + 1

= -(sin4 Î¸ + cos4 Î¸ + 2 sin2 Î¸ cos2 Î¸) + 1

= – (sin2 Î¸ + cos2 Î¸)2 + 1

Also we know that, sin2 Î¸ + cos2 Î¸ = 1

= – 1 + 1

= 0

Then, Right Hand Side = 0

Therefore, LHS = RHS

(vii) sin8 Î¸ – cos8 Î¸ = (sin2 Î¸ – cos2 Î¸)(1 – 2 sin2 Î¸ cos2 Î¸)

From the question first we consider Left Hand Side (LHS),

= sin8 Î¸ – cos8 Î¸

The above terms can be written as,

= (sin4 Î¸)2 – (cos4 Î¸)2

We know that, (a2 – b2) = (a + b)(a – b)

= (sin4 Î¸ – cos4 Î¸) (sin4 Î¸ + cos4 Î¸)

= ((sin2 Î¸)2 – (cos2 Î¸)2) (sin4 Î¸ + cos4 Î¸)

= (sin2 Î¸ – cos2 Î¸) (sin2 Î¸ + cos2 Î¸)(sin4 Î¸ + cos4 Î¸)

= (sin2 Î¸ – cos2 Î¸)(sin4 Î¸ + cos4 Î¸)

= (sin2 Î¸ – cos2 Î¸)((sin2 Î¸)2 + (cos2 Î¸)2 + 2sin2 Î¸ cos2 Î¸ – 2sin2 Î¸ cos2 Î¸)

= (sin2 Î¸ – cos2 Î¸)((sin2 Î¸ + cos2 Î¸)2 – 2 sin2 Î¸ cos2 Î¸)

= (sin2 Î¸ – cos2 Î¸)(1 – 2sin2 Î¸ cos2 Î¸)

Then, Right Hand Side = (sin2 Î¸ – cos2 Î¸)(1 – 2sin2 Î¸ cos2 Î¸)

Therefore, LHS = RHS

(viii) sec4 A – sec2 A = (sin2 A/cos4 A)

From the question first we consider Left Hand Side (LHS),

= sec4 A – sec2 A

We know that, sec A = 1/cos A

= (1/cos4 A) – (1/cos2 A)

= (1 – cos2 A)/cos4 A

Also we know that, 1 – cos2 A = sin2 A

= sin2 A/cos4 A

Then, Right Hand Side = sin2 A/cos4 A

Therefore, LHS = RHS

(ix) (tan2 Î¸/(tan2 Î¸ – 1)) + (cosec2 Î¸/(sec2 Î¸ – cosec2 Î¸)) = (1/(sin2 Î¸ – cos2 Î¸))

From the question first we consider Left Hand Side (LHS),

= (tan2 Î¸/(tan2 Î¸ – 1)) + (cosec2 Î¸/(sec2 Î¸ – cosec2 Î¸))

We know that, tan2 Î¸ = sin2 Î¸/cos2 Î¸, cosec2 Î¸ = 1/sin2 Î¸, sec2 Î¸ = 1/cos2 Î¸

= ((sin2 Î¸/cos2 Î¸)/((sin2 Î¸/cos2 Î¸) – 1)) + ((1/sin2 Î¸)/((1/cos2 Î¸) – (1/sin2 Î¸)))

= (sin2 Î¸/(sin2 Î¸ – cos2 Î¸)) + ((1/sin2 Î¸)/((sin2 Î¸ – cos2 Î¸)/(cos2 Î¸ sin2 Î¸)))

= (sin2 Î¸/(sin2 Î¸ – cos2 Î¸)) + (cos2 Î¸/(sin2 Î¸ – cos2 Î¸))

= (sin2 Î¸ + cos2 Î¸)/(sin2 Î¸ – cos2 Î¸)

= 1/(sin2 Î¸ – cos2 Î¸)

Then, Right Hand Side = 1/(sin2 Î¸ – cos2 Î¸)

Therefore, LHS = RHS

(x) (sec2 Î¸ – sin2 Î¸)/(tan2 Î¸) = cosec2 Î¸ – cos2 Î¸

From the question first we consider Left Hand Side (LHS),

= (sec2 Î¸ – sin2 Î¸)/(tan2 Î¸)

We know that, sec2 Î¸ = 1/cos2 Î¸, tan2 Î¸ = sin2 Î¸/cos2 Î¸

= ((1/cos2 Î¸) – sin2 Î¸)/(sin2 Î¸/cos2 Î¸)

= ((1 – sin2 Î¸ cos2 Î¸)/cos2 Î¸)/(sin2 Î¸/cos2 Î¸)

= (1 – sin2 Î¸ cos2 Î¸)/sin2 Î¸

= (1/sin2 Î¸) – ((sin2 Î¸ cos2 Î¸)/sin2 Î¸)

= cosec2 Î¸ – cos2 Î¸

Then, Right Hand Side = cosec2 Î¸ – cos2 Î¸

Therefore, LHS = RHS

(xi) ((cos3 Î¸ + sin3 Î¸)/(cos θ + sin θ)) + ((cos3 Î¸ – sin3 Î¸)/(cos θ – sin θ)) = 2

From the question first we consider Left Hand Side (LHS),

= ((cos3 Î¸ + sin3 Î¸)/(cos θ + sin θ)) + ((cos3 Î¸ – sin3 Î¸)/(cos θ – sin θ))

By taking LCM we get,

= [((cos3 Î¸ + sin3 Î¸)(cos θ – sin θ)) + ((cos3 Î¸ – sin3 Î¸)(cos θ + sin θ))]/((cos θ + sin θ)(cos θ – sin θ))

Then,

= (cos4 Î¸ – cos3 Î¸ sin θ + sin3 Î¸ cos θ – sin4 Î¸ + cos4 Î¸ + cos3 Î¸ sin θ – sin3 Î¸ cos θ – sin4 Î¸)/(cos2 Î¸ – sin2 Î¸)

By simplification we get,

= (2 cos4 Î¸ – 2 sin4 Î¸)/(cos2 Î¸ – sin2 Î¸)

Taking common outside,

= 2(cos4 Î¸ – sin4 Î¸)/(cos2 Î¸ – sin2 Î¸)

= (2(cos2 Î¸ + sin2 Î¸)(cos2 Î¸ – sin2 Î¸))/(cos2 Î¸ – sin2 Î¸)

= 2(cos2 Î¸ + sin2 Î¸)

We know that, sin2 Î¸ + cos2 Î¸ = 1

= 2

Then, Right Hand Side = 2

Therefore, LHS = RHS

(xii) (tan θ + sin θ)/(tan θ – sin θ) = (sec θ + 1)/(sec θ – 1)

From the question first we consider Left Hand Side (LHS),

= (tan θ + sin θ)/(tan θ – sin θ)

We know that, tan θ = sin θ/cos θ

= ((sin θ/cos θ) + sin θ)/((sin θ/cos θ) – sin θ)

= (sin θ + sin θ cos θ)/(sin θ – sin θ cos θ)

= (sin θ(1 + cos θ))/(sin θ(1 – cos θ))

= (1 + cos θ)/(1 – cos θ)

Also we know that, cos θ = 1/sec θ

= (1 + (1/cos θ))/(1 – (1/sec θ))

= ((sec θ + 1)/sec θ)/((sec θ – 1)sec θ)

= (sec θ + 1)/(sec θ – 1)

Then, Right Hand Side = (sec θ + 1)/(sec θ – 1)

Therefore, LHS = RHS


7. If m = a sec A + b tan A and n = a tan A + b sec A, prove that m2– n2= a2 – b2.

Answer

From the question it is given that,

m = a sec A + b tan A

n = a tan A + b sec A

We have to prove that, m2 – n2 = a2 – b2

Then,

m2 – n2 = (a sec A + b tan A)2 – (a tan A + b sec A)2

We know that, (a + b)2 = a2 + b2 + 2ab

= a2sec2A + b2 tan2 A + 2a sec A b tan A – (a2 tan2 A + b2 sec2 A + 2ab sec A tan A)

= sec2 A(a2 – b2) + tan2 A(b2 – a2)

= (a2 – b2)[sec2 A – tan2 A]

Also we know that, sec2 A – tan2 A = 1

= (a2 – b2)

Hence it is proved that, m2 – n2 = a2 – b2


8. If x = r sin A cos B, y = r sin A sin B and z = r cos A, prove that x2+ y2+ z2 = r2.

Answer

From the question it is given that,

x = r sin A cos B

y = r sin A sin B

z = r cos A

We have to prove that, x2 + y2 + z2 = r2

First we consider Left Hand Side (LHS),

= x2 + y2 + z2

= (r sin A cos B)2 + (r sin A sin B)2 + (r cos A)2

= r2 sin2 A cos2 B + r2 sin2 A sin2 B + r2 cos2 A

Taking common terms outside we get,

= r2 sin2 A (cos2 B + sin2 B) + r2 cos2 A

= r2 (sin2 A + cos2 A)

We know that, sin2 A + cos2 A = 1

= r2

Then, Right Hand Side = r2

Therefore, LHS = RHS

Hence it is proved that, x2 + y2 + z2 = r2


9. If sin A + cos A = m and sec A + cosec A = n, prove that n (m2– 1) = 2m

Answer

From the question it is given that,

sin A + cos A = m

sec A + cosec A = n

We have to prove that, n (m2 – 1) = 2m

First we consider Left Hand Side (LHS),

= n(m2 – 1)

= (sec A + cosec A)(( sin A + cos A)2 – 1)

We know that, sec A = 1/cos A, cosec A = 1/sin A

= ((1/cos A) + (1/sin A))[sin2 A + cos2 A + 2 sin A cos A – 1]

= ((cos A + sin A)/(sin A cos A)) (1 + 2 sin A cos A – 1)

= ((cos A + sin A)/(sin A cos A))(2 sin A cos A)

= 2(sin A + cos A)

= 2m

Then, Right Hand Side = 2m

Therefore, LHS = RHS

Hence it is proved that, n (m2 – 1) = 2m


10. If x = a cos θ, y = b cot θ, prove that (a2/x2) – (b2/y2) = 1.

Answer

From the question it is given that,

x = a cos θ

y = b cot θ

We have to prove that, (a2/x2) – (b2/y2) = 1

First we consider Left Hand Side (LHS),

= (a2/x2) – (b2/y2)

= (a2/a2cos2 Î¸) – (b2/b2 cot2 Î¸)

= (1/cos2 Î¸) – (1/cot2 Î¸)

= sec2 Î¸ – tan2 Î¸

We know that, 1 + tan2 Î¸ = sec2 Î¸

= 1

Then, Right Hand Side = 1

Therefore, LHS = RHS

Hence it is proved that, (a2/x2) – (b2/y2) = 1.


11. If sec θ + tan θ = m, sec θ – tan θ = n, prove that mn = 1.

Answer

From the question it is given that,

sec θ + tan θ = m

sec θ – tan θ = n

We have to prove that, mn = 1

First we consider Left Hand Side (LHS),

= mn

= (sec θ + tan θ)( sec θ – tan θ)

We know that, a2 – b2 = (a + b) (a – b)

= sec2 Î¸ – tan2 Î¸

Also we know that, 1 + tan2 Î¸ = sec2 Î¸

= 1

Then, Right Hand Side = 1

Therefore, LHS = RHS

Hence it is proved that, mn = 1.


12. If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2– y2= a2 – b2

Answer

From the question it is given that,

x = a sec θ + b tan θ

y = a tan θ + b sec θ

We have to prove that, x2 – y2 = a2 – b2

First we consider Left Hand Side (LHS),

= x2 – y2

= (a sec θ + b tan θ)2 – (a tan θ + b sec θ)2

We know that, (a + b)2 = a2 + 2ab + b2

= a2 sec2 Î¸ + b2 tan2 Î¸ + 2 ab sec θ tan θ – (a2 tan2 Î¸ + b2 sec2 Î¸ + 2ab sec θ tan θ)

Then,

= sec2 Î¸ (a2 – b2) + tan2 Î¸ (b2 – a2)

= (a2 – b2)[sec2 Î¸ – tan2 Î¸]

Also we know that, sec2 Î¸ – tan2 Î¸ = 1

= a2 – b2

Then, Right Hand Side = a2 – b2

Therefore, LHS = RHS

Hence it is proved that, x2 – y2 = a2 – b2.


13. If tan A + sin A = m and tan A – sin A = n, prove that (m2– n2)2= 16mn

Answer

From the question it is given that,

tan A + sin A = m

tan A – sin A = n

We have to prove that, (m2 – n2)2 = 16mn

First we consider Left Hand Side (LHS),

= (m2 – n2)2

= [(tan A + sin A)2 – (tan A – sin A)2]2

=[{(tan A + sin A) – (tan A – sin A)}{(tan A + sin A) + (tan A – sin A)}]2

By simplification we get,

= [(2 sin A)(2 tan A)]2

= [4 sin A tan A]2

= 16sin2 A tan2 A

Then, Right Hand Side = 16mn

= 16(tan2 A – sin2 A)

= 16((sin2 A/cos2 A) – sin2 A)

= 16 sin2 A ((1 – cos2 A)/cos2 A)

= 16 sin2 A (sin2 A/cos2 A)

= 16 sin2 A tan2 A

Therefore, LHS = RHS

Hence it is proved that, (m2 – n2)2 = 16mn.


14. If sin A + cos A = √2, prove that sin A cos A = ½

Answer

From the question it is given that, sin A + cos A = √2

We have to prove that, sin A cos A = ½

We know that, (sin A + cos A)2 = sin2 A + cos2 A + 2 sin A cos A

So, 2 = 1 + 2 sin A cos A

2 sin A cos A = 1

sin A cos A = ½


15. If a sin2θ + b cos2θ = c and p sin2 Î¸ + q cos2 Î¸ = r, prove that (b – c)(r – p) = (c – a)(q – r).

Answer

From the question it is given that,

a sin2 Î¸ + b cos2 Î¸ = c

p sin2 Î¸ + q cos2 Î¸ = r

We have to prove that, (b – c)(r – p) = (c – a)(q – r)

Consider, LHS = (b – c)(r – p)

= (b – a sin2 Î¸ + b cos2 Î¸)( p sin2 Î¸ + q cos2 Î¸ – p)

= [b(1 – cos2 Î¸) – a sin2 Î¸] [p(sin2 Î¸ – 1) + q cos2 Î¸]

= [(b – a) sin2 Î¸][(q – p) cos2 Î¸]

= (b – a)(q – p)sin2 Î¸ cos2 Î¸

Now consider, RHS = (c – a) (q – r)

= (a sin2 Î¸ + b cos2 Î¸ – a) (q – p sin2 Î¸ – q cos2 Î¸)

= [(b – a) cos2 Î¸][(q – p) sin2 Î¸]

= (b – a)(q – p) sin2 Î¸ cos2 Î¸

Therefore, LHS = RHS

Hence it is proved that, (b – c)(r – p) = (c – a)(q – r).


Exercise 21.2


1. If m = a sec A + b tan A and n = a tan A + b sec A, prove that m2 – n2 = a2 – b2

Answer:


2. If x = r sin A cos B, y = r sin A sin B and z = r cos A, prove that x2 + y2 + z2 = r2

Answer


3. If sin A + cos A = m and sec A + cosec A = n, prove that n(m2 – 1) = 2m.

Answer


4. If x = a cos θ, y = b cot θ, prove that a2/x2 – b2/y2 = 1.

Answer


5. If sec θ + tan θ, sec θ – tan θ = n, prove that mn = 1

Answer


6. If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 – y2 = a2 – b2

Answer


7. If tan A + sin A = m and tan A – sin A, prove that (m2 – n2)2 = 16 mn

Answer


8. If sin A + cos A = √2, prove that sin A cos A = ½

Answer


9. If a sin2 θ + b cos2 θ = c and p sin2 θ + q cos2 θ = r, prove that (b – c)(r – p) = (c – a)(q – r)

Answer


10. If x/(a cos θ) = y/(b sin θ) and ax/(cos θ) – by/(sin θ) = a2 – b2, prove that x2/a2 + y2/b2 = 1

Answer


Exercise 21.3


1. Without using trigonometric tables, evaluate:

(i) cosec 40° cos 41° +( tan 31°/cot 59°)

(ii) (sin 47°)/(cos 43°) – 4 cos2 45° + (cos 43°)/(sin 47°)2

(iii) cos 90° + sin 30° tan 45° cos2 45°

(iv) {(sin 49°)/(sin 41°)}2 + {(cos 41°)/(sin 49°)}2

(v) (sin 72°)/(cos 18°) – (sec 32°)/(cosec 58°)

Answer


2. Without using trigonometric identities, show that:

(i) sin 42° sec 48° + cos 42° cosec 48° = 2

(ii) tan 10° tan 20° tan 30° tan 70° tan 80° = 1/√3

(iii) sin (50° + θ) – cos (40° - θ) = 0

(iv) cos2 25° + cos2 65° = 1

(v) sec 70° sin 20° - cos 20° cosec 70° = 0

Answer


3. Prove that sin A/(sin 90° - A) + cos A/(cos 90° - A) = {sec (90° - A)cosec (90° - A)}

Answer


4. For ΔABC, prove that:

(i) tan (B + C)/2 = cot A/2

(ii) sin (A + B)/2 = cos C/2

Answer


5. Prove that sin (90° - A). cos (90° - A) = tan A/(1 + tan2 A)

Answer:


6. Find the value of x, if cos x = cos 60° cos 30° - sin 60° sin 30°

Answer:


7. Find x, if cos (2x – 6) = cos2 30° cos2 60°

Answer


8. Given cos 38° sec(90° - 2A)= 1, find the value of ∠A.

Answer


9. Prove that {1/(1 + cos (90° - A)} + {1/(1 – cos (90° - A)} = 2 cosec2 (90° - A)

Answer


10. Find the value of θ (0° < θ < 90°) if :

(i) cos 63° sec (90° - θ) = 1

(ii) tan 35° cot (90° - θ) = 1

Answer

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