Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Important Questions which are most likely to be asked in the exam.

## Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Important Questions

Question 1.
Find the values of
(i) sin $$\frac{5 \pi}{3}$$
(ii) tan (855°)
(iii) sec $$\left(\frac{13 \pi}{3}\right)$$
Solution:
i) sin $$\frac{5 \pi}{3}$$ = sin $$\left(2 \pi-\frac{\pi}{3}\right)$$ = -sin $$\frac{\pi}{3}$$ = $$-\frac{\sqrt{3}}{2}$$

ii) tan (855°) = tan (2 × 360° + 135°)
= tan (135°)
= tan (180° – 45°)
= -tan 45° = -1

iii) sec $$\left(13 \frac{\pi}{3}\right)$$ = sec $$\left(4 \pi+\frac{\pi}{3}\right)$$
= sec $$\frac{\pi}{3}$$ = 2

Question 2.
Simplify.
i) Cot (θ – $$\frac{13 \pi}{2}$$)
ii) tan $$\left(-23 \frac{\pi}{3}\right)$$
Solution:

Question 3.
Find the value of sin2 $$\frac{\pi}{10}$$ + sin2 $$\frac{4 \pi}{10}$$ + sin2 $$\frac{6 \pi}{10}$$ + sin2 $$\frac{9 \pi}{10}$$
Solution:

Question 4.
If sin θ = $$\frac{4}{5}$$ and θ is not in the first qua-drant, find the value of cos θ.
Solution:
∵ sin θ = $$\frac{4}{5}$$ and θ is not in the first quadrant
⇒ θ lies in 2nd quadrant, ∵ sin θ is +ve

Question 5.
If sec θ + tan θ = $$\frac{2}{3}$$, find the value of sin θ and determine the quadrant in which θ lies.
Solution:
∵ sec θ + tan θ = $$\frac{2}{3}$$

∵ tan θ is negative, sec θ is positive
⇒ θ lies in fourth quadrant.

Question 6.
Prove that
cot$$\frac{\pi}{16}$$.cot$$\frac{2 \pi}{16}$$.cot$$\frac{3 \pi}{16}$$…..cot$$\frac{7 \pi}{16}$$ = 1
Solution:

Question 7.
If 3 sin θ + 4 cos θ = 5, then find the value of 4 sin θ – 3 cos θ.
Solution:
Given 3 sin θ + 4 cos θ = 5 and let 4 sin θ – 3 cos θ = x
By squaring and adding, we get
(3 sin θ + 4 cos θ)2 + (4 sin θ – 3 cos θ)2 = 52 + x2
⇒ 9 sin2 θ + 16 cos2 θ + 24 sin θ cos θ + 16 sin2 θ + 9 cos2 θ – 24 sin θ cos θ = 25 + x2
⇒ 9 + 16 = 25 + x2
⇒ x2 = 0 ⇒ x = 0
∴ 4 sin θ – 3 cos θ = 0.

Question 8.
If cos θ + sin θ = $$\sqrt{2}$$ cos θ, prove that cos θ – sin θ = $$\sqrt{2}$$ sin θ. (May ’11)
Solution:
Given cos θ + sin θ = $$\sqrt{2}$$ cos θ
⇒ sin θ = ($$\sqrt{2}$$ – 1) cos θ
Now multiplying both sides by ($$\sqrt{2}$$ + 1)

Question 9.
Find the value of 2(sin2 θ + cos6 θ) – 3(sin4 θ + cos4 θ)
Solution:
2 (sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ)
= 2[(sin2 θ)3 + (cos2 θ)3] – 3[(sin2 θ)2 + (cos2 θ)2]
= 2[(sin2 θ + cos2 θ)3 – 3 sin2 θ cos2 θ (sin2 θ + cos2 θ)] – 3[(sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ]
= 2[1 – 3 sin2 θ cos2 θ] – 3[1 – 2 sin2 θ cos2 θ]
= 2 – 6 sin2 θ cos2 θ – 3 + 6 sin2 θ cos2 θ
= -1

Question 10.
Prove that (tan θ + cot θ)2 = sec2 θ + cosec2 θ = sec2 θ cosec2 θ.
Solution:
(tan θ + cot θ)2
=tan2 θ + cot 2 θ + 2 tan θ cot θ
= tan2 θ + cot2 θ + 2
= (1 + tan2 θ) + (1 + cot2 θ)
= sec2 θ + cosec2 θ

Question 11.
If cos θ > 0, tan θ + sin θ = m and tan θ – sin θ = n, then show that m2 – n2 = n, then show that m2 – n2 = 4$$\sqrt{m n}$$
Solution:
Given that m = tan θ + sin θ
n = tan θ – sin θ
⇒ m + n = 2 tan θ and m – n = 2 sin θ
Now(m + n)(m – n) = 4 tan θ sin θ

Question 12.
If tan 20° = λ, then show that $$\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \cdot \tan 110^{\circ}}$$ = $$\frac{1-\lambda^{2}}{2 \lambda}$$ (A.P.) Mar. ’16
Solution:
Given tan 20° = λ

Question 13.
Find the values of sin 75°, cos 75°, tan 75° and cot 75°.
Solution:
i) sin 75° = sin (45° + 30°)
= sin 45°. cos 30° + cos 45°. sin 30°

ii) cos (75°) = cos (45° + 30°)
= cos 45° cos 30° – sin 45° sin 30°

Question 14.
If 0 < A, B < 90°.cos A = $$\frac{5}{13}$$ and sin B = $$\frac{4}{5}$$-then find sin(A + B).
Solution:
0 < A < 90°and cos A = $$\frac{5}{13}$$ ⇒ sin A = $$\frac{12}{13}$$
0 < B < 90°and sin B = $$\frac{4}{5}$$ ⇒ cos B = $$\frac{3}{5}$$
∴ sin (A + B) = sin A cos B + cos A sin B
= $$\frac{12}{13}$$ • $$\frac{3}{5}$$ + $$\frac{5}{13}$$ • $$\frac{4}{5}$$ = $$\frac{56}{65}$$

Question 15.
Prove that
sin2 $$\left(52 \frac{1}{2}\right)^{\circ}$$ – sin2 $$\left(22 \frac{1}{2}\right)^{\circ}$$ = $$\frac{\sqrt{3}+1}{4 \sqrt{2}}$$
Solution:

Question 16.
Prove that tan 70° – tan 20° = 2 tan 50°.
Solution:
tan 50° = tan (70° – 20°)
= $$\frac{\tan 70^{\circ}-\tan 20^{\circ}}{1+\tan 20^{\circ} \cdot \tan 70^{\circ}}$$
⇒ tan 70° – tan 20°
= tan 50° [1 + tan 20° . tan (90° – 20°)]
= tan 70° – tan 20° = tan 50°[1 + tan 20° cot 20°]
⇒ tan 70° – tan 20° = 2 tan 50°.

Question 17.
If A + B = π/4, then prove that
i)(1 + tan A)(1 + tan B) = 2. (May ’11; Mar. ’07)
ii) (cot A – 1)(cot B – 1) = 2
Solution:
i) Given that A + B = π/4 (T.S) (Mar. ’16)
⇒ tan (A + B) = tan (π/4)
⇒ $$\frac{\tan A+\tan B}{1-\tan A \tan B}$$ = 1
⇒ tan A + tan B = 1 – tan A tan B
⇒ tan A + tan B + tan A tan B = 1
⇒ 1 + tan A + tan B + tan A tan B = 1 + 1
⇒ (1 + tan A)(1 + tan B) = 2

ii) Given A + B = π/4
⇒ cot (A + B) = cot π/4
⇒ $$\frac{\cot A \cot B-1}{\cot B+\cot A}$$ = 1
⇒ cot A cot B – 1 = çot B + cot A
⇒ cot A cot B – cot A – cot B = 1
Again add 1 on both sides
cot A cot B – cot A – cot B + 1 = 1 + 1
∴ (cot A – 1) (cot B – 1) = 2.

Question 18.
If sin α = $$\frac{1}{\sqrt{10}}$$, sin β = $$\frac{1}{\sqrt{5}}$$ and α, β are acute, show that α + β = π/4.
Solution:
Given α is acute and sin α = $$\frac{1}{\sqrt{10}}$$
⇒ tan α = $$\frac{1}{3}$$
β is acute and sin β = $$\frac{1}{\sqrt{5}}$$ ⇒ tan β = $$\frac{1}{2} .$$

Question 19.
If sin A = $$\frac{12}{13}$$, cos B = $$\frac{3}{5}$$ and neither A nor B is in the first quadrant, then find the quadrant in which A + B lies.
Solution:
sin A = $$\frac{12}{13}$$ and A is not in first quadrant
⇒ A lies in second quadrant, ∵ sin A is +ve
cos B = $$\frac{3}{5}$$ and B is not in first quadrant
⇒ B lies in fourth quadrant, cos B is +ve
∵ sin A = $$\frac{12}{13}$$ ⇒ cos A = –$$\frac{5}{13}$$
cos B = $$\frac{3}{5}$$ ⇒ sin B = –$$\frac{4}{5}$$
sin (A + B) = sin A cos B + cos A sin B

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

∵ sin (A + B) and cos (A + B) are positive
⇒ (A + B) lies in first quadrant.

Question 20.
Find (i) tan $$\left(\frac{\pi}{4}+\mathbf{A}\right)$$ interms of tan A
and (ii) cot $$\left(\frac{\pi}{4}+\mathbf{A}\right)$$ interms of cot A.
Solution:

Question 21.
Prove that $$\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}$$ = cot 36°. (A.P) (Mar ’15, Mar. ’11)
Solution:

Question 22.
Show that
cos 42° + cos 78° + cos 162° = 0 (May. ’11)
Solution:
L.H.S. = cos 42° + cos 78° + cos 162°
= 2 cos $$\left(\frac{42^{\circ}+78^{\circ}}{2}\right)$$. cos $$\left(\frac{42^{\circ}-78^{\circ}}{2}\right)$$ + cos (180° – 18°)
= 2 cos 60° . cos (-18°) + cos (180° – 18°)
= 2$$\left(\frac{1}{2}\right)$$ cos 18° – cos 18° = 0
∴ cos 42° + cos 78° + cos 162° = 0

Question 23.
Express $$\sqrt{3}$$ sin θ + cos θ as a sine of an angle.
Solution:
$$\sqrt{3}$$ sin θ + cos θ = 2($$\frac{\sqrt{3}}{2}$$ sin θ + cos θ)
= 2(cos $$\frac{\pi}{6}$$ sin θ + sin $$\frac{\pi}{6}$$ cos θ)
= 2. sin[θ + $$\frac{\pi}{6}$$]

Question 24.
Prove that sin2 θ + sin2 [θ – $$\frac{\pi}{3}$$] + sin2 [θ – $$\frac{\pi}{3}$$] = $$\frac{3}{2} .$$
Solution:

Question 25.
If A, B, C are angles of a triangle and if none of them is equal to $$\frac{\pi}{2}$$, then prove that
i) tan A + tan B + tan C = tan A tan B tan C
ii) cot A cot B + cot B cot C + cot C cot A = 1
Solution:
i) ∵ A, B, C are angles of a triangle
⇒ A + B + C = 180°
⇒ A + B = 180° – C
⇒ tan (A + B) = tan (180° – C)
⇒ $$\frac{\tan A+\tan B}{1-\tan A \tan B}$$ = -tan C
⇒ tan A + tan B = -tan C + tan A tan Btan C
⇒ tan A + tan B + tan C = tan A tan B tan C

ii) ∵ A + B + C = 180°
⇒ A + B = 180° – C
⇒ cot (A + B) = cot (180° – C)
⇒ $$\frac{\cot A \cot B-1}{\cot B+\cot A}$$ = -cot C
⇒ cot A cot B – 1 = -cot B cot C – cot C. cot A
⇒ cot A cot B + cot B cot C + cot C cot A = 1

Question 26.
Let ABC be a triangle such that cot A + cot B + cot C = $$\sqrt{3}$$. Then prove that ABC is an equilateral triangle.
Solution:
Given that A + B + C = 180°
Then we know that
cot A cot B + cot B cot C + cot C cot A = 1
i.e., Σ(cot A cot B) = 1 —– (1)
Now Σ (cot A – cot B)2
= Σ (cot2 A + cot2 B – 2 cot A cot B)
= 2 cot2 A + 2 cot2 B + 2 cot2 C – 2 cot A cot B – 2 cot B cot C – 2 cot C cot A
= 2 [cot A + cot B + cot C]2 – 6 (cot A cot B + cot B cot C + cot C cot A)
= 2
= 6 – 6 = 0
∴ Σ (cot A – cot B)2 = 0 .
⇒ cot A = cot B = cot C
⇒ cot A = cot B = cot C = $$\frac{\sqrt{3}}{3}$$ = $$\frac{1}{\sqrt{3}}$$
(∵ cot A + cot B + cot C = $$\sqrt{3})$$)
⇒ A = B = C = 60°
∴ ΔABC is an equilateral triangle.

Question 27.
Suppose x = tan A, y = tan B, z = tan C. Suppose none of A, B, C, A – B, B – C, C – A is an odd multiple of $$\frac{\pi}{2}$$.Then prove that $$\sum\left(\frac{x-y}{1+x y}\right)$$ = $$\pi\left(\frac{x-y}{1+x y}\right)$$
Solution:
∵ x = tan A, y = tan B, z = tan C

Write P = A – B, Q = B – C, R = C – A.
Then P + Q + R = O
⇒ tan (P + Q) = tan (-R)
⇒ $$\frac{\tan P+\tan Q}{1-\tan P \tan Q}$$ = -tan R
⇒ tan P + tan Q = -tan R + tan P tan Q tan R
⇒ tan P + tan Q + tan R = tan P tan Q tan R
⇒ Σ(tan P) = π (tan P)
⇒ Σtan (A – B) = π tan (A – B)
∴ Σ$$\left(\frac{x-y}{1+x y}\right)$$ = π$$\left(\frac{x-y}{1+x y}\right)$$

Question 28.
Find the values of
i) sin 22$$\frac{1}{2}$$°
ii) cos 22$$\frac{1}{2}$$°
iii) tan 22$$\frac{1}{2}$$°
iv) cot 22$$\frac{1}{2}$$°
Solution:

Question 29.
Find the values of
i) sin 67$$\frac{1}{2}$$°
ii) cos 67$$\frac{1}{2}$$°
iii) tan 67$$\frac{1}{2}$$°
iv) cot 67$$\frac{1}{2}$$°
Solution:
Let A = 67$$\frac{1}{2}$$° ⇒ = 2A = 135°

Question 30.
Simplify: $$\frac{1-\cos 2 \theta}{\sin 2 \theta}$$
Solution:
$$\frac{1-\cos 2 \theta}{\sin 2 \theta}$$ = $$\frac{2 \sin ^{2} \theta}{2 \sin \theta \cos \theta}$$ = $$\frac{\sin \theta}{\cos \theta}$$ = tan θ

Question 31.
If cos A = $$\sqrt{\frac{\sqrt{2}+1}{2 \sqrt{2}}}$$, find the value of cos 2A.
Solution:
cos 2A = 2 cos2A – 1 = 2 $$\left(\frac{\sqrt{2}+1}{2 \sqrt{2}}\right)$$ – 1
= $$\frac{\sqrt{2}+1}{\sqrt{2}}$$ – 1 = $$\frac{1}{\sqrt{2}}$$

Question 32.
If cos θ = $$\frac{-5}{13}$$ and $$\frac{\pi}{2}$$ < θ < π, find the value of sin 2θ.
Solution:
$$\frac{\pi}{2}$$ < θ < π ⇒ sin θ > 0 and cos θ = –$$\frac{5}{13}$$
⇒ Sin θ = $$\frac{12}{13}$$
∴ sin 2θ = 2 sin θ cos θ
= 2 . $$\frac{12}{13}$$(-$$\frac{5}{13}$$) = –$$\frac{120}{169}$$

Question 33.
For what values of x in the first quadrant $$\frac{2 \tan x}{1-\tan ^{2} x}$$ is positive?
Solution:
$$\frac{2 \tan x}{1-\tan ^{2} x}$$ > 0 ⇒ tan 2x > 0
⇒ 0 < 2x < $$\frac{\pi}{2}$$ (since x is in the first quadrant)
⇒ 0 < x < $$\frac{\pi}{4}$$

Question 34.
If cos θ = $$\frac{-3}{5}$$ and π < θ < $$\frac{3 \pi}{2}$$, find the value of tan θ/2.
Solution:

Question 35.
If A is not an integral multiple of $$\frac{\pi}{2}$$, prove that
i) tan A + cot A = 2 cosec 2A
ii) cot A – tan A = 2 cot 2A
Solution:
i) tan A + cot A = $$\frac{\sin A}{\cos A}$$ + $$\frac{\cos A}{\sin A}$$
= $$\frac{\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}$$

Question 36.
If A is not an integral multiple of prove that
i) tan A + cot A = 2 cosec 2A
ii) cot A – tan A = 2 cot 2A
Solution:

Question 37.
If θ is not an integral multiple of $$\frac{\pi}{2}$$, prove that tan θ + 2 tan 2θ + 4 tan 4θ +
8 cot 8θ = cot θ.
Solution:
From cot A – tan A = 2 cot 2A above tan A = cot A – 2 cot 2A ….(1)
∴ tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ = (cot θ – 2 cot 2θ) + 2 (cot 2θ – 2 cot 4θ) + 4 (cot 4θ – 2 cot 8θ) + 8 cot 8θ (by (1) above)
= cot θ

Question 38.
For A ∈ R, prove that
i) sin A.sin(π/3 +A).sin(π/3 – A) = $$\frac{1}{4}$$ . sin3A
ii) cosA . cos (π/3 + A). cos(π/3 – A)
= $$\frac{1}{4}$$ cos 3A and hence deduce that
iii) sin 20° sin 40° sin 60° sin 80° = $$\frac{3}{16}$$
iv) cos $$\frac{\pi}{9}$$, cos $$\frac{2 \pi}{9}$$, cos $$\frac{3 \pi}{9}$$. cos $$\frac{4 \pi}{9}$$ = $$\frac{1}{16}$$.
Solution:
i) sin A. sin (π/3 + A). sin (π/3 – A)
= sin A [sin2 π/3 – sin2 A]

iii) ∵ sin A. sin (60° + A). sin (60° – A)
= $$\frac{1}{4}$$ sin 3A
Put A = 20°
⇒ sin 20°. sin (60° + 20°). sin (60° – 20°)
= $$\frac{1}{4}$$. sin(3 × 20°) .
⇒ sin 20°. sin 40°. sin 80° = $$\frac{1}{4}$$ sin 60°
Multiplying on both sides with sin 60°
We get, sin 20° sin 40° sin 60° sin 80°
= $$\frac{1}{4}$$ sin2 60°
= $$\frac{1}{4}\left(\frac{\sqrt{3}}{2}\right)^{2}[latex] = [latex]\frac{3}{16}$$

iv) ∵ cos A. cos (60° + A). cos (60° – A) = $$\frac{1}{4}$$cos 3A
Put A = 20°
⇒ cos 20°. cos (60° + 20°) cos (60° – 20°)
= $$\frac{1}{4}$$. cos(3 × 20°)
⇒ cos 20°. cos 40°. cos 80° = cos 60°
On multiplying both sides by cos 60°, we get cos 20°. cos 40°. cos 60°. cos 80°
= $$\frac{1}{4}$$. cos2 60°

Question 39.
If 3A is not an odd multiple of $$\frac{\pi}{2}$$, prove that tan A. tan (60° + A). tan (60° – A)
= tan 3A and hence find the value of tan 6°. tan 42°. tan 66°. tan 78°.
Solution:

∴ tan A . tan $$\left(\frac{\pi}{3}+\mathrm{A}\right)$$ tan $$\left(\frac{\pi}{3}-A\right)$$ = tan 3A
i.e., tan A tan (60° – A) tan (60° + A) = tan 3A —— (1)
Put A = 6°
⇒ tan 6° tan 54° tan 66° = tan 18° ——- (2)
Put A = 18° in (1)
tan 18° tan 42° tan 78° = tan 54° ——- (3)
put (2) in (3),
(tan 6° tan 54° tan 66°) tan 42° tan 78°
= tan 54
⇒ tan 6° tan 42° tan 66° tan 78° = 1

Question 40.
For α, β ∈ R, prove that (cos α + cos β)2 + (sin α + sin β)2 = 4 cos2 $$\left(\frac{\alpha-\beta}{2}\right)$$.
Solution:
LH.S. = (cos α + cos β)2 + (sin α + sin β)2
= (cos2 α + cos2 β + 2 cos α cos β + (sin2 α + sin2 β + 2 sin α sin β)
= 1 + 1 + 2 (cos α cos β + sin α sin β)
= 2 + 2 cos (α – β) = 2 (1 + cos (α – β)]
= 2[2cos2$$\left(\frac{\alpha-\beta}{2}\right)$$]
= 4.cos2$$\left(\frac{\alpha-\beta}{2}\right)$$; [∵ 1 + cos θ = 2 cos2 $$\frac{\theta}{2}$$]

Question 43.
If a, b, c are non zero real numbers and α, β are solutions of the equation a cos θ + b sin θ = c then show that
(i) sin α + sin β = $$\frac{2 b c}{a^{2}+b^{2}}$$ and
(ii) sin α. sin β = $$\frac{c^{2}-a^{2}}{a^{2}+b^{2}}$$
Solution:
∵ a cos θ + b sin θ = c
= a cos θ = c – b sin θ
⇒ (a cos θ)2 = (c – b sin θ)2
⇒ a2 cos2 θ = c2 + b2 sin2 θ – 2bc sinθ
⇒ a2 (1 – sin2 θ) = c2 + b2 sin2 θ – 2bc sin θ
⇒ (a2 + b2) sin2 θ – 2bc sin θ + (c2 – a2) = 0
This is a quadratic equatioñ in sin θ, whose roots are sin α and sin β (given that α, β are two solutions of θ)
∴ Sum of the roots sin α + sin β = $$\frac{2 b c}{a^{2}+b^{2}}$$
Product of the roots sin α sin β = $$\frac{c^{2}-a^{2}}{a^{2}+b^{2}}$$

Question 42.
If θ is not an odd multiple of $$\frac{\pi}{2}$$ and cos θ ≠ $$\frac{-1}{2}$$, prove that
$$\frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}$$ = tan θ.
Solution:

Question 43.
Prove that sin4 $$\frac{\pi}{8}$$ + sin4 $$\frac{3 \pi}{8}$$ + sin4 $$\frac{5 \pi}{8}$$ + sin4 $$\frac{7 \pi}{8}$$ = $$\frac{3}{2}$$
Solution:

Question 44.
If none of 2A and 3A is an odd multiple of $$\frac{\pi}{2}$$, then prove that tan 3A. tan 2A. tanA = tan 3A – tan 2A – tan A.
Solution:
∵ tan 3A = tan (2A + A)
= $$\frac{\tan 2 A+\tan A}{1-\tan 2 A \tan A}$$
⇒ tan 3A(1 – tan 2A tan A) = tan 2A + tan A
⇒ tan 3A – tan A tan 2A tan 3A
= tan 2A + tan A
∴ tan 3A – tan 2A – tan A
= tan A tan 2A tan 3A

Question 45.
Prove that sin 78° + cos 132° = $$\frac{\sqrt{5}-1}{4}$$.
Solution:
L.H.S. = sin 78° + cos 132°
= sin 78° + cos (90° + 42°)
= sin 78° – sin 42°

Question 46.
Prove that sin 21° cos 9° – cos 84° cos 6° = $$\frac{1}{4}$$.
Solution:
LH.S. = sin 21° cos 9° – cos 84° cos 6°
= $$\frac{1}{2}$$[2 sin 21° cos 9° – 2 cos 84° cos 6°]
= $$\frac{1}{2}$$[sin (21° + 9°) + sin (21° – 9°) – 2 cos (90° – 6°) cos 6°]
= $$\frac{1}{2}$$[sin 30° + sin 12° – 2 sin 6° cos 6°]
= $$\frac{1}{2}$$[$$\frac{1}{2}$$ + sin 12° – sin (2 × 16°)
= $$\frac{1}{4}$$ = R.H.S.

Question 47.
Find the value of sin 34° + cos 64° – cos 4°.
Solution:
sin 34° + (cos 64° – cos 4°)

Question 48.
Prove that cos2 76° + cos2 16° – cos 76° cos 16° = $$\frac{3}{4} .$$
Solution:
L.H.S. = cos2 76° + cos2 16° – cos 76° cos 16°
= cos2 76° + (1 – sin2 16°) – $$\frac{1}{2}$$(2 cos 76° cos 16°)
= 1 + (cos2 76° – sin2 16°) – $$\frac{1}{2}$$[cos (76° + 16°) + cos (76° – 16°)]
= 1 + cos (76° + 16°). cos (76° – 16°) – $$\frac{1}{2}$$[cos 92° + cos 60°]
= 1 + cos (92°). cos 60° – $$\frac{1}{2}$$cos 92° – $$\frac{1}{2}$$cos 60°
= 1 + $$\frac{1}{2}$$. cos 92° – $$\frac{1}{2}$$. cos 92° – $$\frac{1}{2}$$ × $$\frac{1}{2}$$
= 1 – $$\frac{1}{4}$$ = $$\frac{3}{4}$$ = R.H.S.

Question 49.
If a, b, ≠ 0 and sin x + sin y = a and cos x + cos y = b, find two values of
i) tan $$\left(\frac{x+y}{2}\right)$$
ii) sin $$\left(\frac{x-y}{.2}\right)$$ interms of a and b.
Solution:

First method:

ii) a2 + b2 = (sin x + sin y)2 + (cos x + cos y)2
= sin2 x + sin2 y + 2 sin x sin y + cos2 x + cos2 y + 2 cos x cos y
= 2 + 2(cos x cos y + sin x sin y)
= 2 + 2 cos (x – y)
= 2[1 + cos (x – y)]
a2 + b2 – 2 = 2 cos (x – y)

Second method:

Question 50.
Prove that cos 12° + cos 84° + cos 132° + cos 156° = –$$\frac{1}{2}$$.
Solution:
L.H.S. = cos 12° + cos 84° + cos 132° + cos 156°

Question 51.
Show that for any θ ∈ R
4 sin $$\frac{5 \theta}{2}$$ cos $$\frac{3 \theta}{2}$$ cos 3θ = sin θ – sin 2θ + sin 4θ + sin 7θ
Solution:
R.H.S. = 4 sin $$\frac{5 \theta}{2}$$ cos $$\frac{3 \theta}{2}$$ cos 3θ

= 2 [(sin 4θ + sin θ) cos 3θ]
= 2 sin 4θ cos 3θ + 2 sinθ cos 3θ
= sin (4θ + 3θ) + sin (4θ – 3θ) + sin (θ + 3θ) + sin(θ – 3θ)
= sin 7θ + sin θ + sin 4θ – sin 2θ = R.H.S.

Question 52.
If none of A, B, A + B is an integral multiple of π, then prove that

Solution:

Question 53.
For any α ∈ R, provethat cos2 (α – π/4) + cos2 (α + π/12) – cos2 (α – π/12) = $$\frac{1}{2}$$.
Solution:

Question 54.
Suppose (α – β) is not an odd multiple of $$\frac{\pi}{2}$$, m is a non – zero real number such that m ≠ -1 and $$\frac{\sin (\alpha+\beta)}{\cos (\alpha-\beta)}$$ = $$\frac{1-m}{1+m}$$, Then prove that tan $$\left(\frac{\pi}{4}-\alpha\right)$$ = m. tan $$\left(\frac{\pi}{4}+\beta\right)$$
Solution:

Question 55.
If A, B, C are the angles of a triangle, prove that
i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
ii) sin 2A + sin 2B – sin 2C = 4 cos A cos B sin C
Solution:
i) ∵ A, B, C are angles of a triangle
⇒ A + B + C = 180° ——- (1)
L.H.S. = sin 2A + sin 2B + sin 2C

= 2 sin (A + B) cos (A – B) + sin 2C
= 2 sin (180° – C) cos (A – B) + sin 2C
By(1)
= 2 sin C. cos (A — B) + 2 sin C. cos C
= 2 sin C [cos (A — B) + cos C]
= 2 sin C [cos (A — B) + cos (180° — $$\overline{A+B}$$)]
= 2 sin C [cos (A – B) — cos (A + B)]
= 2 sin C (2 sin A sin B)
= 4 sin A sin B sin C = R.H.S.

ii) LH.S. = sin 2A + sin 28 — sin 2C

= 2 sin(A+ B)cos(A—B)—sin2C
= 2 sin (180° — C) cos (A — B) — sin 2C
= 2 sin C. cos(A – B) – 2 sin C cos C
= 2 sin C [cos (A – B) – cos C]
= 2 sin C [cos (A — B)—cos (180°— $$\overline{A+B}$$)]
By(1)
= 2 sin C [cos (A — B) + cos (A + B)]
= 2 sin C [2 cos A cas B]
= 4 cos A cos B sin C = R.H.S.

Question 56.
If A, B, C are angles of a triangle, prove that
i) cos 2A + cos 2B + cos 2C
= -4 cos A cos B cos C – 1

ii) cos 2A + cos 2B – cos 2C
= 1 – 4 sin A sin B cos C
Solution:
i) ∵ A, B, C are angles of a triangle
⇒ A + B + C = 180° ———— (1)
L.H.S. = (cos 2A + cos 2B) + cos 2C
= 2 cos(A + B) cos (A – B) + 2 cos2 C – 1
= -2 cos C cos (A – B) – 2 cos C cos (A + B) – 1
[∵ cos (A + B) = -cos C]
= – 1 – 2 cos C[cos (A – B) + cos(A – B)]
= – 1 – 2 cos C[2 cos A cos B]
= -1 – 4 cos A cos B cos C = R.H.S.

ii) L.H.S. = cos 2A + cos 2B – cos 2C

= 2 cos (A + B). cos (A – B) – (2 cos2 C – 1)
= 2 cos (180° – C) cos (A – B) – 2 cos2 C + 1
= 1 – 2 cos C cos (A – B) – 2 cos2 C
= 1 – 2 cos C [cos (A – B) + cos C]
= 1 – 2 cos C[cos(A – B) + cos(180° – $$\overline{\mathrm{A}+\mathrm{B}}$$)]
= 1 – 2 cos C [cos (A – B) – cos (A + B)]
= 1 – 2 cos C [2 sin A sin B]
= 1 – 4 sin A sin B cos C = R.H.S.

Question 57.
If A B, C are angles in a triangle, then prove that
i) sin A + sin B + sin C = 4 cos $$\frac{A}{2}$$ cos $$\frac{B}{2}$$ cos $$\frac{C}{2}$$
ii) cos A + cos B + cos C = 1 + 4 sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$ sin $$\frac{C}{2}$$
Solution:

Question 58.
If A + B + C = π/2, then show that
i) sin2 A + sin2 B + sin2 C = 1 – 2 sin A sin B sin C
ii) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
Solution:
A + B + C = π/2 ——– (1)
L.H.S. = sin2 A + sin2 B + sin2 C
= $$\frac{1}{2}$$(1 – cos2A + 1 – cos 2B + 1 – cos 2C]
= $$\frac{1}{2}$$[3 – (cos 2A + cos 2B + cos 2C)]
= $$\frac{1}{2}$$[3 – (1 + 4 sin A sin B sin C)
(By Problem No. 57(ii)]
(∵ 2A + 2B + 2C = 180°)
= $$\frac{1}{2}$$[2 – 4 sin A sin B sin C]
= 1 – 2 sin A sin B sin C

ii) A + B + C = 90°
⇒ 2A + 2B + 2C = 180°
2A + 2B + 2C 180°
sin 2A + sin 2B + sin 2C
= 4 cos A cos B cos C
By Problem No. 57(i).

Question 59.
If A + B + C = 3π/2, prove that cos 2A + cos 2B + cos 2C = 1 – 4 sin A sin B sin C.
Solution:
A + B + C = 3π/2 —— (1)
L.H.S. = cos 2A + cos 2B + cos 2C
= 2 cos (A + B). cos(A – B) +1 – 2 sin2 C
= 2 cos (270° – C). cos (A – B) + 1 – 2 sin2 C
= 1 – 2 sin C cos (A – B) – 2 sin2C
= 1 – 2 sin C [cos(A – B) + sin C]
= 1 – 2 sin C[cos (A – B) + sin (270° – $$\overline{A+B}$$)]
= 1 – 2 sin C[cos(A – B) – cos(A + B)]
= 1 – 2 sin C[2 sin A sin B]
= 1 – 4 sin A sin B sin C

Question 60.
If A B, C are angles of a triangle, then prove that
sin2 $$\frac{A}{2}$$ + sin2 $$\frac{B}{2}$$ – sin2 $$\frac{C}{2}$$ = 1 – 2 cos $$\frac{A}{2}$$ cos $$\frac{B}{2}$$ sin $$\frac{C}{2}$$ (Mar. ’16, May ’12, ’11)
Solution:

Question 61
If A. B, C are angles of a triangle, then prove that sin $$\frac{\mathbf{A}}{\mathbf{2}}$$ + sin $$\frac{\mathbf{B}}{\mathbf{2}}$$ + sin $$\frac{\mathbf{C}}{\mathbf{2}}$$ = 1 + 4 sin $$\frac{\pi-\mathbf{A}}{4}$$ sin $$\frac{\pi-B}{4}$$ sin $$\frac{\mathrm{C}}{2}$$
Solution:
A + B + C = 180° ——- (1)

Question 62.
If A + B + C = 0, then prove that cos2 A + cos2 B + cos2 C = 1 + 2 cos A cos B cos C.
Solution:
A + B + C = 0 —— (1)
L.H.S. = cos2 A + cos2 B + cos2 C
= cos2 A + (1 – sin2 B) + cos2 C
= 1 + (cos2 A – sin2 B) + cos2 C
= 1 + cos (A + B) cos (A – B) + cos2 C
= 1 + cos(-C) cos(A – B) + cos2 C
By (1)
= 1 + cos C cos (A – B) + cos2 C
= 1 + cos C [cos (A – B) + cos c]
= 1 + cos C[cos(A – B) + cos(-B – A)]
By (1)
1 + cos C[cos (A – B) + cos(A + B)]
= 1 + cos C (2 cos A cos B)
= 1 + 2 cos A cos B cos C = R.H.S.

Question 63.
If A + B + C = 25, then prove that cos (S — A) + cos (S — B) + cos (S — C) + cos (S) = 4 cos $$\frac{A}{2}$$ cos $$\frac{B}{2}$$ cos $$\frac{C}{2}$$.
Solution: