ICSE Solutions for Selina Concise Chapter 21 Trigonometrical Identities Class 10 Maths
Exercise 21(A)
Prove the following identities:
1. (sec A – 1)/(sec A + 1) = (1 – cos A)/(1 + cos A)
Solution
Hence, Proved
2. (1 + sin A)/(1 – sin A) = (cosec A + 1)/(cosec A – 1)
Solution
Hence, Proved
3. 1/(tan A + cot A) = cos A sin A
Solution
Taking L.H.S,
Hence, Proved
4. tan A – cot A = (1 – 2 cos2 A)/(sin A cos A)
Solution
Taking LHS,
Hence, Proved
5. sin4 A – cos4 A = 2 sin2 A – 1
Solution
Taking L.H.S,
sin4 A – cos4 A
= (sin2 A)2 – (cos2 A)2
= (sin2 A + cos2 A) (sin2 A – cos2 A)
= sin2A – cos2A
= sin2A – (1 – sin2A) [Since, cos2 A = 1 – sin2 A]
= 2sin2 A – 1
– Hence Proved
6. (1 – tan A)2 + (1 + tan A)2 = 2 sec2 A
Solution
Taking L.H.S,
(1 – tan A)2 + (1 + tan A)2
= (1 + tan2 A + 2 tan A) + (1 + tan2 A – 2 tan A)
= 2 (1 + tan2 A)
= 2 sec2 A [Since, 1 + tan2 A = sec2 A]
Hence, Proved
7. cosec4 A – cosec2 A = cot4 A + cot2 A
Solution
cosec4 A – cosec2 A
= cosec2 A(cosec2 A – 1)
= (1 + cot2 A) (1 + cot2 A – 1)
= (1 + cot2 A) cot2 A
= cot4 A + cot2 A = R.H.S
Hence, Proved
8. sec A (1 – sin A) (sec A + tan A) = 1
Solution
Taking L.H.S,
sec A (1 – sin A) (sec A + tan A)
Hence, Proved
9. cosec A (1 + cos A) (cosec A – cot A) = 1
Solution
Taking L.H.S,
Hence, Proved
10. sec2 A + cosec2 A = sec2 A . cosec2 A
Solution
Taking L.H.S,
Hence, Proved
11. (1 + tan2 A) cot A/ cosec2 A = tan A
Solution
Taking L.H.S,
= RHS
Hence, Proved
12. tan2 A – sin2 A = tan2 A. sin2 A
Solution
Taking L.H.S,
tan2 A – sin2 A
Hence, Proved
13. cot2 A – cos2 A = cos2A. cot2A
Solution
Taking L.H.S,
cot2 A – cos2 A
Hence, Proved
14. (cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
Solution
Taking L.H.S,
(cosec A + sin A) (cosec A – sin A)
= cosec2 A – sin2 A
= (1 + cot2 A) – (1 – cos2 A)
= cot2 A + cos2 A = R.H.S
Hence, Proved
15. (sec A – cos A)(sec A + cos A) = sin2 A + tan2 A
Solution
Taking L.H.S,
(sec A – cos A)(sec A + cos A)
= (sec2 A – cos2 A)
= (1 + tan2 A) – (1 – sin2 A)
= sin2 A + tan2 A = RHS
Hence, Proved
16. (cos A + sin A)2 + (cosA – sin A)2 = 2
Solution
Taking L.H.S,
(cos A + sin A)2 + (cosA – sin A)2
= cos2 A + sin2 A + 2cos A sin A + cos2 A – 2cosA.sinA
= 2 (cos2 A + sin2 A) = 2 = R.H.S
Hence, Proved
17. (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1
Solution
Taking LHS,
(cosec A – sin A)(sec A – cos A)(tan A + cot A)
= RHS
Hence, Proved
18. 1/ sec A + tan A = sec A – tan A
Solution
Taking LHS,
= RHS
Hence, Proved
19. cosec A + cot A = 1/ cosec A – cot A
Solution
Taking LHS,
cosec A + cot A
= RHS
Hence, Proved
20. sec A – tan A/sec A + tan A = 1 – 2 secA tanA + 2 tan2 A
Solution
Taking LHS,
= 1 + tan2 A + tan2 A – 2 sec A tan A
= 1 – 2 sec A tan A + 2 tan2 A = RHS
Hence, Proved
21. (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Solution
Taking LHS,
(sin A + cosec A)2 + (cos A + sec A)2
= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2cos A sec A
= (sin2 A + cos2 A ) + cosec2 A + sec2 A + 2 + 2
= 1 + cosec2 A + sec2 A + 4
= 5 + (1 + cot2 A) + (1 + tan2 A)
= 7 + tan2 A + cot2 A = RHS
Hence, Proved
22. sec2 A. cosec2 A = tan2 A + cot2 A + 2
Solution
Taking,
RHS = tan2 A + cot2 A + 2 = tan2 A + cot2 A + 2 tan A. cot A
= (tan A + cot A)2 = (sin A/cos A + cos A/ sin A)2
= (sin2 A + cos2 A/ sin A.cos A)2 = 1/ cos2 A. sin2 A
= sec2 A. cosec2 A = LHS
Hence, Proved
23. 1/1 + cos A + 1/1 – cos A = 2 cosec2 A
Solution
Taking LHS,
= RHS
Hence, Proved
24. 1/1 – sin A + 1/1 + sin A = 2 sec2 A
Solution
Taking LHS,
= RHS
Hence, Proved
Exercise 21(B)
1. Prove that:
(i) cos A/(1 – tan A) + sin A /(1 – cot A) = sin A + cos A
(ii) (cos3 A + sin3 A)/(cos A + sin A) + (cos3 A − sin3 A)/(cos A + sin A) = 2
(iii) tan A/(1 – cot A) + cot A/(1 –tan A) = sec A cosec A +1
(iv) (tan A + 1/cos A)2 + (tan A – 1/cos A)2 = 2 [(1+sin2 A)/(1 – sin2A)]
(v) 2 sin2 A + cos4 A = 1 + sin4 A
(vi) (sin A – sin B)/(cos A + cos B) + (cos A – cos B)/(sin A + sin B) = 0
(vii) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)
(viii) (1+ tan A tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
(ix) 1/(cos A + sin A -1) + 1/(cos A + sin A +1) = cosec A + sec A
Solution
(i)
Hence, Proved
(ii) Taking LHS,
Hence, Proved
(iii)
Hence, Proved
(iv)
Hence, Proved
(v) Taking LHS,
2 sin2 A + cos2 A
= 2 sin2 A + (1 – sin2 A)2
= 2 sin2 A+ 1 + sin4 A – 2 sin2 A
= 1 + sin4 A = RHS
Hence, Proved
(vi)
Hence, Proved
(vii)
Hence, Proved
(viii) (1 + tanA tanB)2 + (tanA – tanB)2
= 1 + tan2A tan2B + 2tanA tanB + tan2A + tan2B – 2tanA tanB
= 1 + tan2A + tan2B + tan2A tan2B
= sec2A + tan2B(1 + tan2A)
= sec2A + tan2B sec2A
= sec2A(1 + tan2B)
= sec2A sec2B
(ix)
Hence, Proved
2. If x cos A + y sin A = m and x sin A – y cos A = n, then prove that:
x2 + y2 = m2 + n2
Solution
Taking RHS,
m2 + n2
= (x cos A + y sin A)2 + (x sin A – y cos A)2
= x2 cos2 A + y2 sin2 A + 2xy cos A sin A + x2 sin2 A + y2 cos2 A – 2xy sin A cos A
= x2 (cos2 A + sin2 A) + y2 (sin2 A + cos2 A)
= x2 + y2 [Since, cos2 A + sin2 A = 1]
= RHS
3. If m = a sec A + b tan A and n = a tan A + b sec A, prove that m2 – n2 = a2 – b2
Solution
Taking LHS,
m2 – n2
= (a sec A + b tan A)2 – (a tan A + b sec A)2
= a2 sec2 A + b2 tan2 A + 2 ab sec A tan A – a2 tan2 A – b2 sec2 A – 2ab tan A sec A
= a2 (sec2 A – tan2 A) + b2 (tan2 A – sec2 A)
= a2 (1) + b2 (-1) [Since, sec2 A – tan2 A = 1]
= a2 – b2
= RHS
Exercise 21(C)
1. Show that:
(i) tan 10o tan 15o tan 75o tan 80o = 1
Solution
Taking, tan 10o tan 15o tan 75o tan 80o
= tan (90o – 80o) tan (90o – 75o) tan 75o tan 80o
= cot 80o cot 75o tan 75o tan 80o
= 1 [Since, tan θ × cot θ = 1]
(ii) sin 42o sec 48o + cos 42o cosec 48o = 2
Solution
Taking, sin 42o sec 48o + cos 42o cosec 48o
= sin 42o sec (90o – 42o) + cos 42o cosec (90o – 42o)
= sin 42o cosec 42o + cos 42o sec 42o
= 1 + 1 [Since, sin θ × cosec θ = 1 and cos θ × sec θ = 1]
= 2
(iii) sin 26o/sec 64o + cos 26o/cosec 64o = 1
Solution
Taking,
2. Express each of the following in terms of angles between 0°and 45°:
(i) sin 59°+ tan 63°
(ii) cosec 68°+ cot 72°
(iii) cos 74°+ sec 67°
Solution
(i) sin 59°+ tan 63°
= sin (90 – 31)°+ tan (90 – 27)°
= cos 31°+ cot 27°
(ii) cosec 68°+ cot 72°
= cosec (90 – 22)°+ cot (90 – 18)°
= sec 22°+ tan 18°
(iii) cos 74°+ sec 67°
= cos (90 – 16)°+ sec (90 – 23)°
= sin 16°+ cosec 23°
3. Show that:
(i) sin A/sin(90° −A) + cosA/cos(90° −A) = sec A cosec A
(ii) sin A cos A− sin Acos (90° −A)cosA/sec(90° −A) – cos Asin(90° −A)sin A/cosec (90° −A) = 0
Solution
= sin A cos A – sin3 A cos A – cos3 A sin A
= sin A cos A – sin A cos A (sin2 A + cos2 A)
= sin A cos A – sin A cos A (1) [Since, sin2 A + cos2 A = 1]
= 0
4. For triangle ABC, show that:
(i) sin (A + B)/2 = cos C/2
(ii) tan (B + C)/2 = cot A/2
Solution
We know that, in triangle ABC
∠A + ∠B + ∠C = 180o [Angle sum property of a triangle]
(i) Now,
(∠A + ∠B)/2 = 90o – ∠C/2
So, sin {(A + B)/2} = sin (90o – C/2)
= cos C/2
(ii) And,
(∠C + ∠B)/2 = 90o – ∠A/2
So,
tan ((B + C)/2) = tan (90o – A/2)
= cot A/2
5. Evaluate:
(i) 3(sin72°/cos 18°) – sec 32°/cosec 58°
(ii) 3cos80° cosec10° + 2 cos59°cosec 31°
(ii) sin80°/cos10° + sin59° sec 31°
(iv) tan(55° - A ) – cot(35° + A)
(v) cosec(65° + A) – sec(25° −A)
(vi) 2(tan57°/cot 33°) – cot70°/tan 20° − √2 cos45°
(vii) cot241°/tan249° −2sin2 75°/cos2 15°
(viii) cos 70°/sin 20° +cos 59°/sin 31° − 8sin2 30°
(ix) 14 sin 30° + 6 cos 60° − 5tan45°
Solution
(i)
(ii) 3 cos 80o cosec 10o + 2 cos 59o cosec 31o
= 3 cos (90 – 10)o cosec 10o + 2 cos (90 – 31)o cosec 31o
= 3 sin 10o cosec 10o + 2 sin 31o cosec 31o
= 3 + 2 = 5
(iii) sin 80o/cos 10o + sin 59o sec 31o
= sin (90 – 10)o/cos 10o + sin (90 – 31)o sec 31o
= cos 10o/cos 10o + cos 31o sec 31o
= 1 + 1 = 2
(iv) tan (55o – A) – cot (35o + A)
= tan [90o – (35o + A)] – cot (35o + A)
= cot (35o + A)] – cot (35o + A)
= 0
(v) cosec (65o + A) – sec (25o – A)
= cosec [90o – (25o – A)] – sec (25o – A)
= sec (25o – A) – sec (25o – A)
= 0
(vi)
(vii)
= 1 – 2 = -1
(viii)
(ix) 14 sin 30o + 6 cos 60o – 5 tan 45o
= 14 (1/2) + 6 (1/2) – 5(1)
= 7 + 3 – 5
= 5
6. A triangle ABC is right angled at B; find the value of (sec A. cosec C – tan A. cot C)/ sin B
Solution
As, ABC is a right angled triangle right angled at B
So, A + C = 90o
(sec A. cosec C – tan A. cot C)/sin B
= (sec (90o – C). cosec C – tan (90o – C). cot C)/sin 90o
= (cosec C. cosec C – cot C. cot C)/1 = cosec2 C – cot2 C
= 1 [Since, cosec2 C – cot2 C = 1]
Exercise 21(D)
1. Use tables to find sine of:
(i) 21°
(ii) 34° 42′
(iii) 47° 32′
(iv) 62° 57′
(v) 10° 20′ + 20° 45′
Solution
(i) sin 21o = 0.3584
(ii) sin 34o 42’= 0.5693
(iii) sin 47o 32’
= sin (47o 30′ + 2′)
=0.7373 + 0.0004
= 0.7377
(iv) sin 62o 57′
= sin (62o 54′ + 3′)
= 0.8902 + 0.0004
= 0.8906
(v) sin (10o 20′ + 20o 45′)
= sin 30o65′
= sin 31o5′
= 0.5150 + 0.0012
= 0.5162
2. Use tables to find cosine of:
(i) 2° 4’
(ii) 8° 12’
(iii) 26° 32’
(iv) 65° 41’
(v) 9° 23’ + 15° 54’
Solution
(i) cos 2° 4’ = 0.9994 – 0.0001 = 0.9993
(ii) cos 8° 12’ = cos 0.9898
(iii) cos 26° 32’ = cos (26° 30’ + 2’) = 0.8949 – 0.0003 = 0.8946
(iv) cos 65° 41’ = cos (65° 36’ + 5’) = 0.4131 -0.0013 = 0.4118
(v) cos (9° 23’ + 15° 54’) = cos 24° 77’ = cos 25° 17’ = cos (25° 12’ + 5’) = 0.9048 – 0.0006 = 0.9042
3. Use trigonometrical tables to find tangent of:
(i) 37°
(ii) 42° 18′
(iii) 17° 27′
Solution
(i) tan 37o = 0.7536
(ii) tan 42o 18′ = 0.9099
(iii) tan 17o 27′ = tan (17o 24′ + 3′) = 0.3134 + 0.0010 = 0.3144
Exercise 21(E)
1. Prove the following identities:
(i) 1/(cos A + sin A) + 1/(cos A – sin A) = 2cos A/(2cos2A – 1)
(ii) cosec A – cot A = sin A /(1+ cos A)
(iii) 1- sin2A/(1+ cos A) = cos A
(iv) (1- cos A)/sin A + sin A/ 1- cos A = 2 cosec A
(v) cot A /1 – tan A + tan A/ (1-cot A) = 1+ tan A + cot A
(vi) cos A/ (1+ sin A)+ tan A = sec A
(vii) sin A/(1-cos A) – cot A = cosec A
(viii) (sin A –cos A+1)/(sin A + cos A – 1) = cos A/(1-sin A)
(ix)√[(1+sin A)/(1 – sin A)] = cos A/ (1 – sinA)
(x) √[(1 – cos A)/(1+cos A) = sin A/(1 + cos A)
(xi) 1+(sec A – tan A)2 /cosec A(sec A – tan A) = 2 tan A
(xii) [(cosec A – cot A)2 + 1]/sec A(cosec A – cot A) = 2 cot A
(xiii) cot2 A{(sec A – 1 )/(1+ sin A)} + sec2 A{(sin A – 1 )/(1+ sec A)} = 0
(xiv) {(1- 2sin2 A)2/(cos 4A – sin4A) = 2 cos2 A – 1
(xv) sec4 A(1 –sin4 A)−2tan2A = 1
(xvi) cosec4A(1−cos4A)− 2cot2 A = 1
(xvii) (1+tan A + sec A)(1+ cotA – cosec A) = 2
Solution
(i) Taking LHS,
1/(cos A + sin A) + 1/(cos A – sin A)
= RHS
Hence, Proved
(ii) Taking LHS, cosec A – cot A
= RHS
Hence, Proved
(iii) Taking LHS, 1 – sin2 A/(1 + cos A)
= RHS
Hence. Proved
(iv) Taking LHS,
(1 – cos A)/sin A + sin A/(1 – cos A)
= RHS
Hence, Proved
(v) Taking LHS, cot A/(1 – tan A) + tan A/(1 – cot A)
= RHS
Hence, Proved
(vi) Taking LHS, cos A/ (1 + sin A) + tan A
= RHS
Hence, Proved
(vii) Consider LHS,
= (sin A/(1 – cos A)) – cot A
We know that, cot A = cos A/sin A
So,
= (sin2 A – cos A + cos2 A)/(1 – cos A) sin A
= (1–cos A)/(1–cos A) sin A
= 1/sin A
= cosec A
(viii) Taking LHS, (sin A – cos A + 1)/ (sin A + cos A – 1)
= RHS
Hence Proved
(ix) Taking LHS,
= RHS
Hence, Proved
(x) Taking LHS,
= RHS
Hence, Proved
(xi) Taking LHS,
= RHS
Hence, Proved
(xii) Taking LHS,
= RHS
Hence, Proved
(xiii) Taking LHS,
= RHS
Hence, Proved
(xiv) Taking LHS,
= RHS
Hence, Proved
(xv) Taking LHS,
sec4 A (1 – sin4 A) – 2 tan2 A
= sec4 A(1 – sin2 A) (1 + sin2 A) – 2 tan2 A
= sec4 A(cos2 A) (1 + sin2 A) – 2 tan2 A
= sec2 A + sin2 A/ cos2 A – 2 tan2 A
= sec2 A – tan2 A
= 1 = RHS
Hence, Proved
(xvi) cosec4 A(1 – cos4 A) – 2 cot2 A
= cosec4 A (1 – cos2 A) (1 + cos2 A) – 2 cot2 A
= cosec4 A (sin2 A) (1 + cos2 A) – 2 cot2 A
= cosec2 A (1 + cos2 A) – 2 cot2 A
= cosec2 A + cos2 A/sin2 A – 2 cot2 A
= cosec2 A + cot2 A – 2 cot2 A
= cosec2 A – cot2 A
= 1 = RHS
Hence, Proved
(xvii) (1 + tan A + sec A) (1 + cot A – cosec A)
= 1 + cot A – cosec A + tan A + 1 – sec A + sec A + cosec A – cosec A sec A
= 2 + cos A/sin A+ sin A/cos A – 1/(sin A cos A)
= 2 + (cos2 A + sin2 A)/sin A cos A – 1/(sin A cos A)
= 2 + 1/(sin A cos A) – 1/(sin A cos A)
= 2 = RHS
Hence, Proved
2. If sin A + cos A = p and sec A + cosec A = q, then prove that: q(p2 – 1) = 2p
Solution
Taking the LHS, we have
q(p2 – 1) = (sec A + cosec A) [(sin A + cos A)2 – 1]
= (sec A + cosec A) (sin2 A + cos2 A + 2 sin A cos A – 1)
= (sec A + cosec A) (1 + 2 sin A cos A – 1)
= (sec A + cosec A) (2 sin A cos A)
= 2sin A + 2 cos A
= 2p
3. If x = a cos θ and y = b cot θ, show that: a2/x2 – b2/y2 = 1
Solution
Taking LHS,
a2/x2 – b2/y2
4. If sec A + tan A = p, show that: sin A = (p2 – 1)/ (p2 + 1)
Solution
Taking RHS, (p2 – 1)/(p2 + 1)
5. If tan A = n tan B and sin A = m sin B, prove that:
cos2 A = m2 – 1/n2 – 1
Solution
Given,
tan A = n tan B
n = tan A/ tan B
And, sin A = m sin B
m = sin A/ sin B
Now, taking RHS and substitute for m and n
m2 – 1/n2 – 1
6. (i) If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
(ii) If 4 cos2 A – 3 = 0, show that: cos 3A = 4 cos2 A – 3 cos A
Solution
(i) Given, 2 sin A – 1 = 0
So, sin A = ½
We know, sin 30o = 1/2
Hence, A = 30o
Now, taking LHS
sin 3A = sin 3(30o) = sin 30o = 1
RHS = 3 sin 30o – 4 sin3 30o = 3 (1/2) – 4 (1/2)3 = 3 – 4(1/8) = 3/2 – ½ = 1
∴ LHS = RHS
(ii) Given, 4 cos2 A – 3 = 0
4 cos2 A = 3
⇒ cos2 A = 3/4
⇒ cos A = √3/2
We know, cos 30o = √3/2
Hence, A = 30o
Now, taking
LHS = cos 3A = cos 3(30o) = cos 90o = 0
RHS = 4 cos3 A – 3 cos A = 4 cos3 30o – 3 cos 30o = 4 (√3/2)3 – 3 (√3/2)
= 4 (3√3/8) – 3√3/2
= 3√3/2 – 3√3/2
= 0
∴ LHS = RHS
7. Evaluate:
(i) 2(tan 35°/cot55°)2 + (cot 55°/tan 35°)2 –(sec 40°/cosec 50°)
(ii)sec 26° sin64° + cosec 33°/sec 57°
(iii) 5 sin66°/cos24° − 2cot 85°/tan5°
(iv) cos 40° cosec 50° + sin50° sec40°
(v) sin27° sin63° − cos63° cos27°
(vi) 3 sin72°/cos18° −sec32°/cosec58°
(vii) 3 cos 80° cosec10° + 2 cos59°cosec31°
(viii) cos75°/sin15° + sin12°/cos78° − cos18°/sin72°
Solution
(i)
= 2 (1)2 + 12 – 3
= 2 + 1 – 3 = 0
(ii)
= 1 + 1 = 2
(iii)
(iv) cos 40o cosec 50o + sin 50o sec 40o
= cos (90 – 50)o cosec 50o + sin (90 – 50)o sec 40o
= sin 50o cosec 50o + cos 40o sec 40o
= 1 + 1 = 2
(v) sin 27o sin 63o – cos 63o cos 27o
= sin (90 – 63)o sin 63o – cos 63o cos (90 – 63)o
= cos 63o sin 63o – cos 63o sin 63o
= 0
(vi)
(vii) 3 cos 80o cosec 10o + 2 cos 59o cosec 31o
= 3 cos (90 – 10)o cosec 10o + 2 cos (90 – 31)o cosec 31o
= 3 sin 10o cosec 10o + 2 sin 31o cosec 31o
= 3 + 2 = 5
(viii)
8. Prove that:
(i) tan (55o + x) = cot (35o – x)
(ii) sec (70o – θ) = cosec (20o + θ)
(iii) sin (28o + A) = cos (62o – A)
(iv) 1/(1 + cos (90o – A)) + 1/(1 – cos (90o – A)) = 2 cosec2 (90o – A)
(v) 1/(1 + sin (90o – A)) + 1/(1 – sin (90o – A)) = 2 sec2 (90o – A)
Solution:
(i) tan (55o + x) = tan [90o – (35o – x)] = cot (35o – x)(ii) sec (70o – θ) = sec [90o – (20o + θ)] = cosec (20o + θ)
(iii) sin (28o + A) = sin [90o – (62o – A)] = cos (62o – A)
(iv)
(v)
