# RS Aggarwal Solutions Chapter 5 Trigonometric Ratios Exercise -5 Class 10 Maths

 Chapter Name RS Aggarwal Chapter 5 Trigonometric Ratios Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 5 Related Study NCERT Solutions for Class 10 Maths

### Exercise 5 Solutions

1. If sin θ = √3/2, find the value of all T- ratios of θ

Solution

Let us first draw a right ∆ABC, right angled at B and ∠C = θ

Now, we know that sin θ = Perpendicular/Hypotenuse

= AB/AC

= √3/2

So, if AB = √3k, then AC = 2k, where k is a positive number.

Now, using Pythagoras theorem, we have:

AC2 = AB2 + BC2

⇒ BC2 = AC2 – AB2 = (2k)2 – (√3k)2

⇒ BC2 = 4k2 – 3k2 = k2

⇒ BC = k

Now, finding the other T-rations using their definitions, we get:

cos θ = BC/AC = k/2k = 1/2

tan θ = AB/BC = √3k/k = √3

∴ cot θ = 1/tan θ = 1/√3,

cosec θ = 1/sin θ = 2/√3

sec θ = 1/cos θ = 2

2. If cos θ = 7/25, find the value of all T-ratios of θ

Solution

Let us first draw a right ∆ABC, right angled at B and ∠C = θ.

Now, we know that cos θ = Base/hypotenuse = BC/AC = 7/25

So, if BC = 7k, then AC = 25k, were k is a positive number.

Now, using Pythagoras theorem, we have:

AC2 = AB2 + BC2

⇒ AB2 = AC2 – BC2 = (25k)2 – (7k)2

⇒ AB2 = 625k2 – 49k2 = 576k2

⇒ AB = 24k

Now, finding the trigonometric ratios using their definitions, we get:

sin θ = AB/AC = 24k/25k = 24/25

tan θ = AB/BC = 24k/7k = 24/7

∴ cot θ = 1/tan θ = 7/24,

cosec θ = 1/sin θ = 25/24

sec θ = 1/cos θ = 25/7

3. If tan θ = 15/8, find the values of all T-ratios of θ

Solution

Let us first draw a right ∆ABC, right angled at B and ∠C = θ

Now, we know that tan θ = Perpendicular/Base = AB/BC = 15/8

So, if BC = 8k, then AB = 15k where k is positive number.

Now, using Pythagoras theorem, we have:

AC2 = AB2 + BC2 = (15k)2 + (8k)2

⇒ AC2 = 225k2 + 64k2 = 289k2

⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:

Sin θ = AB/AC = 15k/17k = 15/17

cos θ = BC/AC = 8k/17k = 8/17

∴ cot θ = 1/tan θ = 8/15,

cosec θ = 1/sin θ = 17/15

sec θ = 1/cos θ = 17/8

4. If cot θ = 2 find all the values of all T-ratios of θ

Solution

Let us first draw a right ∆ABC, right angled at B and ∠C = θ

Now, we know that cot θ = Base/Perpendicular = BC/AB = 2

So, if BC = 2k, then AB = k, is a positive number.

Now, using Pythagoras theorem, we have:

AC2 = AB2 + BC2 = (2k)2 + (k)2

⇒ AC2 = 4k2 + k2 = 5k2

⇒ AC = √5k

Now, finding the other T-ratios using their definitions, we get:

sin θ = AB/AC = 5/√5k = 1/√2

cos θ = BC/AC = 2k/√5k = 2/√5

∴ tan θ = 1/cot θ = 1/2,

cosec θ = 1/sin θ = √5

sec θ = 1/cos θ = √5/2

5. If cosec θ = 10 find all the values of all T-ratios of θ

Solution

Let us first draw a right ∆ABC, right angled at B and ∠C = θ

6. If sin θ = (a2 – b2)/(a2 + b2) find all T-ratios of θ

Solution

We have sin θ = (a2 – b2)/(a2 + b2)

As,

Cos2 θ = 1 − sin2 θ

= 1 – (a2 – b2)/(a2 + b2)2

= 1/1 – (a2 – b2)2/(a2 + b2)2

= (a2 + b2)2 – (a2 – b2)2/(a2 + b2) + (a2 – b2)]/(a2 + b2)2

= [a2 + b2 – a2 + b2][a2 + b2 + a2 – b2]/(a2 + b2)2

= [2b2][2a2]/(a2 + b2)2

⇒ cos2 θ = 4a2b2/(a2 + b2)2

⇒ cos θ = 2ab/(a2 + b2)

Also,

tan θ = sin θ/cos θ

= [(a2 – b2)/(a2 + b2)]/[2ab/(a2 + b2)]

= (a2 – b2)/2ab

Now,

cosec θ = 1/sin θ

= 1/(a2 – b2)/(a2 – b2)

= (a2 + b2)/(a2 – b2)

Also,

sec θ = 1/cos θ

= 1/(2ab/(a2 + b2))

= (a2 + b2)/2ab

And,

cot θ = 1/tan θ

= 1/(a2 – b2)/2ab

= 2ab/(a2 – b2)

7. If 15 cot A = 8 find all the values of sin A and sec A

Solution

We have,

15 cot A = 8

⇒ cot A = 8/15

As,

cosec2 A = 1 + cot2 A

= 1 + (8/15)2

= 1 + 64/225

= (225 + 64)/225

⇒ cosec2A = 289/225

⇒ cosec A = 17/15

1/sin A = 17/15

Sin A = 15/17

Also,

Cos2 A = 1 – sin2 A

= 1 – (15/17)2

= 1 – 225/289

= (289 – 225)/289

⇒ cos2 A = 64/289

⇒ cos A = 8/17

⇒ 1/sec A = 8/17

⇒ sec A = 17/8

8. If sin A = 9/41 find all the values of cos A and tan A

Solution

We have sin A = 9/41

As,

Cos2 A = 1 − sin2

= 1 − (9/41)2

= 1 – 81/1681

= (1681 – 81)/1681

⇒ cos2 A = 1600/1681

⇒ cos A = 40/41

Also,

Tan A = sin A/cos A

= (9/41)/(40/41)

= 9/40

9. If cos θ = 0.6 show that (5sin θ - 3tan θ) = 0

Solution

Let us consider a right ∆ABC right angled at B.

Now, we know that cos θ = 0.6 = BC/AC = 3/5

So, if BC = 3k, then AC = 5k, where k is a positive number.

Using Pythagoras theorem, we have:

AC2 = AB2 + BC2

⇒ AB2 = AC2 – BC2

⇒ AB2 = (5k)2 – (3k)2 = 25k2 – 9k2

⇒ AB2 = 16k2

⇒ AB = 4k

Finding out the other T-rations using their definitions, we get:

Sin θ = AB/AC = 4k/5k = 4/5

tan θ = AB/BC = 4k/3k = 4/3

Substituting the values in the given expression, we get:

5 sin θ − 3 tan θ

⇒ 5(4/5) – 3(4/3)

⇒ 4 – 4 = 0 = RHS

i.e., LHS = RHS

Hence, Proved.

10. If cosec θ = 2 show that (cot θ + (sin θ/1 + cos θ) = 2

Solution

Let us consider a right ∆ABC, right angled at B and ∠C = θ.

Now, it is given that cosec θ = 2.

Also, sin θ = 1/cosec θ = 1/2 = AB/AC

So, if AB = k, then AC = 2k, where k is a positive number.

Using Pythagoras theorem, we have:

⇒ AC2 = AB2 + BC2

⇒ BC2 = AC2 – AB2

⇒ BC2 (2k)2 – (k)2

⇒ BC2 = 3k2

⇒ BC = √3k

Finding out the other T-ratios using their definitions, we get:

cos θ = BC/AC = √3k/2k = √3/2

tan θ = AB/BC = k/√3k = 1/√3

cot θ = 1/tan θ = √3

Substituting these values in the given expression, we get:

cot θ + sin θ/(1 + sin θ)

= (√3 + (1/2))/(1 + √3/2)

= (√3 + [(1/2)/(2 + √3)]/2

= √3 + 1/(2 + √3)

= (√3(2 + √3) + 1)/(2 + √3)

= (2√3 + 3 + 1)/(2 + √3)

= 2(2 + √3)/(2 + √3) = 2

i.e., LHS = RHS

Hence proved.

11. If tan θ = 1/√7 show that (cosec2 θ – sec2 θ)/(cosec2 θ + sec2 θ) = 3/4

Solution

Let us consider a right ∆ABC, right angled at B and ∠C = θ.

Now it is given that tan θ = AB/BC = 1/√7

So, if AB = k, then BC = √7k, where k is a positive number.

Using Pythagoras theorem, we have:

AC2 = AB2 + BC2

⇒ AC2 = (k)2 + (√7k)2

⇒ AC2 = k2 + 7k2

Now, finding out the values of the other trigonometric ratios, we have:

Sin θ = AB/AC = k/2√2k = 1/2√2

Cos θ = BC/AC = √7k/2√2k = √7/2√2

∴ cosec θ = 1/sin θ = 2√2 and sec θ = 1/cos θ = 2√2/√7

Substituting the values of cosec θ and sec θ in the give expression, we get:

(cosec2 θ – sec2 θ)/(cosec2 θ + sec2 θ)

= [(2√2)2 – (2√2)/√7)2]/[(2√2)2 + (2√2) /√7]2

= [8 – (8/7)]/[8 + (8/7)

= [(56 – 8)/7]/[56 + 8)/7]

= 48/64 = 3/4 = RHS

i.e., LHS = RHS

Hence proved.

12. If tan θ = 20/21, show that (1 – sin θ + cos θ)/(1 + sin θ + cos θ) = 3/7

Solution

Let us consider a right ∆ABC right angled at B and ∠C = θ

Now, we know that tan θ = AB/BC = 2θ/21

So, if AB = 20k, then BC = 21k, where k is a positive number.

Using Pythagoras theorem, we get:

AC2 = AB2 + BC2

⇒ AC2 = (20k)2 + (21k)2

⇒ AC2 = 841k2

⇒ AC = 29k

Now, sin θ = AB/AC = 20/29 and cos θ = BC/AC = 21/29

Substituting these values in the give expression, we get:

L.H.S = (1 - sin θ + cos θ)/(1 + sin θ + cos θ)

= [(1 – 20/29 + 21/29)]/[(1 + 20/29 + 21/29)]

= [(29 – 20 + 21|)/29]/[(29 + 20 + 21)/29

= 30/70

= 3/7 = R.H.S.

∴ LHS = RHS

Hence proved.

13. If sec θ = 5/4 show that (sin θ – 2 cos θ)/(tan θ – cot θ) = 12/7

Solution

We have,

Sec θ = 5/4

⇒ 1/cos θ = 5/4

⇒ cos θ = 4/5

Also,

sin2 θ = 1 – cos2 θ

= 1 – (4/5)2

= 1 – 16/25

= 9/25

⇒ sin θ = 3/5

Now,

L/H.S. = (sinθ – 2 cosθ)/(tanθ - cotθ)

= (sinθ – 2 cosθ)/(sin θ/cosθ – cosθ/sinθ)

= (sinθ – 2 cosθ)/(sin2θ – cos2θ)/(sinθ. cosθ)

= sinθ cosθ (sin θ – 2 cosθ)/(sin2θ – cos2θ)

= [3/5 × 4/5(3/5 – 2 × 4/5)]/[(3/5)2 – (4/5)2]

= 12/25(3/5 – 8/5)/(9/25 – 16/25)

= [12/25 × (-5/3)]/(-7)/25

= 12/7

= RHS

14. If cot θ = 3/4, show that = √(secθ-cosecθ)/(secθ+cosecθ= 1/√7

Solution

L.H.S. =

= 1/7
= RHS

15. If sin θ = 3/4, show that = √(cosec2θ-cot2θ)/(sec2θ-1) = √7/3

Solution

L.H.S. =

= √7/3

= RHS

16. If sin θ = a/b, show that (sec θ + tan θ) = √(b+a)/(b-a)

Solution

LHS = (sec θ + tan θ)

= 1/cos θ + sin θ/cos θ

= (1 + sin θ)/cos θ

= RHS

17. If cos θ = 3/5, show that (sin θ – cot θ)/ 2 tan θ = 3/160

Solution

L.H.S. = (sin θ – cot θ)/2 tan θ

= [sin θ – cos θ/sin θ]/[2(sin θ/cos θ)]

= [(sin2 θ – cos θ)/sin θ]/[(2 sin θ/cos θ)]

= [cos θ(sin2 θ – cos θ)/2 sin2 θ](2 sin2 θ]

= [cos θ(1 – cos2 θ – cos θ)]/[2(1 – cos2 θ)]

= 3/5[1 – (3/5)2 – 3/5]/2[1 – (3/5)2]

= [3/5(1/1 – 9/25 – 3/5)]/[2(1 – 9/25)]

= [3/5(25 – 9 – 15)/25]/[2(25 – 9)/25]

= [3/5(1/25)]/[2(16/25)]

= 3/(5 × 2 × 16)

= 3/160

= RHS

18. If tan θ = 4/3, show that (sin θ + cos θ) = 7/5

Solution

Let us consider a right ∆ABC, right angled at B and ∠C = θ

Now, we know that tan θ = AB/BC = 4/3

So, if BC = 3k, then AB = 4k, where k is a positive number.

Using Pythagoras theorem, we have:

AC2 = AB2 + BC2 = (4k)2 + (3k)2

⇒ AC2 = 16k2 + 9k2 = 25k2

⇒ AC = 5k

Finding out the values of sin θ and cos θ using their definitions, we have:

Sin θ = AB/AC = 4k/5k = 4/5

Cos θ = BC/AC = 3k/5k = 3/5

Substituting these values in the given expressions, we get:

(sin θ + cos θ) = (4/5 + 3/5) = (7/5) = RHS

i.e., LHS = RHS

Hence proved.

19. If tan θ = a/b, show that (a sin θ – b cos θ)/(a sin θ + b cos θ) = (a2 – b2)/(a2 + b2)

Solution

It is given that tan θ = a/b

LHS = (a sin θ – b cos θ)/(a sin θ + b cos θ)

Dividing the numerator and denominator by cos θ, we get:

(a tan θ – b)/(a tan θ + b) (∴ tan θ = sin θ/cos θ)

Now, substituting the value of tan θ in the above expression, we get:

[a(a/b) – b]/[a(a/b) + b]

= [a2/b – b]/[a2/b + b]

= (a2 – b2)/(a2 + b2)

= RHS

i.e., LHS = RHS

Hence proved.

20. If 3 tan θ = 4, show that (4 cos θ – sin θ)/(4 cos θ + sin θ) = 4/5

Solution

Let us consider a right ∆ABC right angled at B and ∠C = θ.

We know that tan θ = AB/BC = 4/3

So, if BC = 3k, then AB = 4k, where k is a positive number.

Using Pythagoras theorem, we have:

AC2 = AB2 + BC2

⇒ AC2 = 16k2 + 9k2

⇒ AC2 = 25k2

⇒ AC = 5k

Now, we have:

Sin θ = AB/AC = 4k/5k = 4/5

Cos θ = BC/AC = 3k/5k = 3/5

Substituting these values in the given expression we get:

(4 cos θ – sin θ)/(2 cos θ + sin θ)

= [4(3/5) – 4/5]/[2(3/5) + 4/5)]

= (12/5 + 4/5)/(6/5 + 4/5)

= [(12 – 4)/5]/[6 + 4)/5]

= 8/10

= 4/5

= RHS

i.e., LHS = RHS

Hence proved.

21. If 3cot θ = 2, show that (4 sin θ – 4 cos θ/0/(2 sin θ + 6 cos θ) = 1/3

Solution

It is given that cos θ = 2/3

LHS = (4 sin θ – 3 cos θ)/(2 sin θ + 6 cos θ)

Dividing the above expression by sin θ, we get:

(4 – 3 cos θ)/(2 + 6 cot θ) [∵ cot θ = cos θ/cot θ]

Now, substituting the values of cot θ in the above expression, we get:

= [4 – 3(2/3)]/[2 + 6(2/3)]

= (4 – 2)/(2 + 4)

= 2/6

= 1/3

i.e., LHS = RHS

Hence proved.

22. If 3 cot θ = 4 show that (1 – tan2 θ)/(1 + tan2 θ) = (cos2 θ – sin2 θ)

Solution

LHS = (1 – tan2 θ)(1 + tan2 θ)

= (1 – 1/cot2 θ)/(1 + 1/cot2 θ)

= [(cot2 θ – 1)/cot2 θ]/[(cot2 θ + 1)/cot2 θ]

= (cot2 θ – 1)/(cot2 θ + 1)

= [(4/3)2 – 1]/[(4/3)2 + 1] (As 3 cot θ = 4 or cot θ = 4/3)

= (16/9 – 1)/(16/9 + 1)

= [(16 – 9)/9]/[(16 + 9)/9]

= (7/9)/(25/9)

= 7/25

RHS = (cos2 θ – sin2 θ)

= (cos2 θ – sin2 θ)/1

= [(cos2 θ – sin2 θ)/sin2 θ]/(1/sin2 θ)

= [cos2 θ/sin2 θ – sin2 θ/sin2 θ]/cosec2 θ

= (cot2 θ – 1)/(cot2 θ + 1)

= [(4/3)2 – 1]/[(4/3)2 + 1]

= (16/9 – 1/1)/(16/9 + 1/1)

= (16 – 9)/9/(16 + 9)/9

= (7/9)/(25/9)

= 7/25

Since, LHS = RHS

Hence, verified.

23. If sec θ = 17/8 verify that (3 – 4 sin2 θ)/(4 cos2 θ – 3) = (3 – tan2 θ)/(1 – tan3 θ)

Solution

It is given that sec θ = 17/8

Let us consider a right ∆ABC right angled at B and ∠C = θ

We know that cos θ = 1/sec θ = 8/17 = BC/AC

So, if BC = 8k, then AC = 17k, where k is a positive number.

Using Pythagoras theorem, we have:

AC2 = AB2 + BC2

⇒ AB2 = AC2 – BC2 = (17k)2 – (8k)2

⇒ AB2 = 289k2 – 64k2 = 225k2

⇒ AB = 15k.

Now, tan θ = AB/BC = 15/8 and sin θ = AB/AC = 15k/17k = 15/17

The given expression is (3 – 4 sin2 θ)/(4 cos2 θ – 3) = (3 – tan2 θ)/(1 – 3 tan2 θ)

Substituting the values in the above expression, we get:

LHS = [3 – 4(15/17)2]/[4(8/17)2 – 3]

= (3 – 900/289)/(256/289 – 3)

= (867 – 900)/(256 – 867)

= -(33/-611)

= 33/611

RHS = [3 – (15/8)2]/[(1 – 3(15/8)2]

= [3 – 225/64]/[1 – 675/64]

= (192 – 255)/(64 – 675)

= - (33/-611) = 33/611

∴ LHS = RHS

Hence proved.

24. In the adjoining figure, B = 90°, BAC = θ°, BC = CD = 4 cm and AD = 10 cm. find (i) sin θ and (ii) cos θ

Solution

In ∆ABD,

Using Pythagoras theorem, we get

Again,

In ∆ABC,

Using Pythagoras theorem, we get

Now,

(i) sin θ = BC/AC

(ii) cos θ = AB/AC

25. In a ΔABC, B = 90°, AB = 24 cm and BC = 7 cm find (i) sin A (ii) cos A (iii) sin C (iv) cos C

Solution

AC2 = AB2 + BC2

⇒ AC2 = (24)2 + (7)2

⇒ AC2 = 576 + 49 = 625

⇒ AC = 25 cm

Now, for T-Ratios of ∠A, base = AB and perpendicular = BC

(i) sin A = BC/AC = 7/25

(ii) cos A = AB/AC = 24/25

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB

(iii) sin C = AB/AC = 24/25

(iv) cos C = BC/AC = 7/25

26. In ΔABC, ∠C = 90°, ∠ABC = θ° BC = 21 units and AB = 29 units. Show that (cos2 θ – sin2 θ) = 41/841

Solution

Using Pythagoras theorem, we get:

AB2 = AC2 + BC2

⇒ AC2 = AB2 – BC2

⇒ AC2 = (29)2 – (21)2

⇒ AC2 = 841 – 441

⇒ AC2 = 400

= 20 units

Now, sin θ = AC/AB = 2θ/29 and cos θ = BC/AB = 21/29

cos2 θ – sin2 θ = (21/29)2 – (20/29)2

= 441/841 – 400/841

= 41/841

Hence proved.

27. In a ΔABC, ∠B = 90°, AB = 12 cm and BC = 5 cm Find (i) cos A (ii) cosec A (iii) cos C (iv) cosec C

Solution

Using Pythagoras theorem, we get:

AC2 = AB2 + BC2

Now, for T-Ratios of ∠A, base = AB and perpendicular BC

(i) cos A = AB/AC = 12/13

(ii) cosec A = 1/sin A = AC/BC = 13/5

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB

(ii) cos C = BC/AC = 5/13

(iv) cosec C = 1/sin C = AC/AB = 13/12

28. If sin α = 1/2 prove that (3 cos α – 4 cos2 α) = 0

Solution

LHS = (3cos α − 4 cos3 α)

= cos α (3 − 4 cos2 α)

= 0

= RHS

29. If ΔABC, ∠B = 90° ABD Tan A = 1/√3. Prove that

(i) Sin A. cos C + cos A. Sin c = 1

(ii) cos A. cos C - sin A. sin C = 0

Solution

In ∆ABC, ∠B = 900

As, tan A = 1/√3

⇒ BC/AB = 1/√3

Let BC = x and AB = x√3

Using Pythagoras we get

Now,

(i) LHS = sin A. cos C + cos A. sin C

= (BC/AC)2 + (AB/AC)2

= (x/2x)2 + (x√3/2x)2

= 1/4 + 3/4

= 1

= RHS

(ii) LHS = cos A. cos C – sin A. sin C

= AB/AC.BC/AC – BC/AC.AB/AC

= (x√3/2x.x/2x) – (x/2x.x√3/2x)

= √3/4 - √3/4

= 0

= RHS

30. If √A and √B are acute angles such that Sin A = Sin B prove that A = B

Solution

In ∆ABC, √C = 90°

Sin A = BC/AB and

Sin B = AC/AB

As, sin A = sin B

⇒ BC/AB = AC/AB

⇒ BC = AC

So, ∠A = ∠B (Angles opposite to equal sides are equal)

31. If ∠A and ∠B are acute angles such that tan A = Tan B then prove that ∠A = ∠B

Solution

In ∆ABC, ∠C = 900

Tan A = BC/AC and

Tan B = AC/BC

As tan A = tan B

⇒ BC/AC = AC/BC

⇒ BC2 = AC2

⇒ BC = AC

So, ∠A = ∠B (Angles opposite to equal sides are equal)

32. If a right ΔABC, right-angled at B, if tan A = 1 then verify that 2sin A. cos A = 1

Solution

We have,

Tan A = 1

⇒ sin A/cos A = 1

⇒ sin A = cos A

⇒ sin A – cos A = 0

Squaring both sides,

(sin A – cos A)2 = 0

⇒ sin2 A + cos2 A – 2 sin A. cos A = 0

⇒ 1 – 2 sin A. cos A = 0

∴ 2 sin A. cos A = 1

33. In the figure of ΔPQR, ∠P = θ° P θ and ∠R = φ° find

(i) √(x+1) cotɸ

(ii) √(x3 +x2) cotɸ

(iii) cos θ

Solution

In ∆PQR, ∠Q = 900,

Using Pythagoras theorem, we get

Now,

(i)

(ii)

(iii) cos θ

34. If x = cosec A +cos A and y = cosec A – cos A then prove that (2/x + y)2 + (x – y/2)2 = 1

Solution

LHS = (2/x + y)2 + (x – y/2)2 – 1

= [2/(cosec A + cos A) + (cosec A – cos A)]2 + [(cosec A + cos A) – (cosec A – cos A)/2]2 – 1

= [2/(cosec A + cos A + cosec A – cos A]2 + [cosec A + cos A – cosec A + cos A/2]2 – 1

= [2/2 cosec A]2 + [2 cos A/2]2 – 1

= [1/cosec A]2 + [cos A]2 – 1

= [sin A]2 + [cos A]2 – 1

= sin2 A + cos2 A - 1

= 1 - 1

= RHS

35. If x = cot A + cos A and y = cot A – cos A then prove that [(x – y)/(x + y)]2 + (x – y/2)2 = 1

Solution

LHS = (x – y/x + y)2 + (x – y/2)2

= [(cot A + cos A – cot A – cos A)/(cot A + cos A + cot A – cos A) + [(cot A + cos A – cot A + cos A)/2]2

= [2cos A/2cot A]2 + [2cos A/2]2

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

= [sin A cos A/cos A]2 + [cos A]2

= sin2 A + cos2 A

= 1

= RHS