# Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry – II Miscellaneous Exercise 3

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Miscellaneous Exercise 3 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Miscellaneous Exercise 3

I. Select the correct option from the given alternatives.

Question 1.
The value of sin(n + 1) A sin(n + 2) A + cos(n + 1) A cos(n + 2) A is equal to
(a) sin A
(b) cos A
(c) -cos A
(d) sin 2A
(b) cos A
Hint:
L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos [(n + 1)A] . cos [(n + 2)A]
= cos [(n + 2)A] . cos [(n + 1)A] + sin [(n + 2)A] . sin [(n + 1)A]
Let (n + 2)A = a and (n + 1)A = b … (i)
∴ L.H.S. = cos a . cos b + sin a . sin b
= cos (a – b)
= cos [(n + 2)A – (n + 1)A] ……..[From (i)]
= cos [(n + 2 – n – 1)A]
= cos A
= R.H.S.

Question 2.
If tan A – tan B = x and cot B – cot A = y, then cot (A – B) = ________
(a) $$\frac{1}{y}-\frac{1}{x}$$
(b) $$\frac{1}{x}-\frac{1}{y}$$
(c) $$\frac{1}{x}+\frac{1}{y}$$
(d) $$\frac{x y}{x-y}$$
(c) $$\frac{1}{x}+\frac{1}{y}$$
Hint:

Question 3.
If sin θ = n sin(θ + 2α), then tan(θ + α) is equal to
(a) $$\frac{1+n}{2-n}$$ tan α
(b) $$\frac{1-n}{1+n}$$ tan α
(c) tan α
(d) $$\frac{1+n}{1-n}$$ tan α
(d) $$\frac{1+n}{1-n}$$ tan α
Hint:

Question 4.
The value of $$\frac{\cos \theta}{1+\sin \theta}$$ is equal to ________
(a) $$\tan \left(\frac{\theta}{2}-\frac{\pi}{4}\right)$$
(b) $$\tan \left(-\frac{\pi}{4}-\frac{\theta}{2}\right)$$
(c) $$\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$$
(d) $$\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)$$
(c) $$\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$$
Hint:

Question 5.
The value of cos A cos (60° – A) cos (60° + A) is equal to ________
(a) $$\frac{1}{2}$$ cos 3A
(b) cos 3A
(c) $$\frac{1}{4}$$ cos 3A
(d) 4cos 3A
(c) $$\frac{1}{4}$$ cos 3A
Hint:

Question 6.
The value of $$\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}$$ is ________
(a) $$\frac{1}{16}$$
(b) $$\frac{1}{64}$$
(c) $$\frac{1}{128}$$
(d) $$\frac{1}{256}$$
(b) $$\frac{1}{64}$$
Hint:

Question 7.
If α + β + γ = π, then the value of sin2 α + sin2 β – sin2 γ is equal to ________
(a) 2 sin α
(b) 2 sin α cos β sin γ
(c) 2 sin α sin β cos γ
(d) 2 sin α sin β sin γ
(c) 2 sin α sin β cos γ
Hint:
sin2 α + sin2 β – sin2 γ
= $$\frac{1-\cos 2 \alpha}{2}+\frac{1-\cos 2 \beta}{2}-\sin ^{2} \gamma$$
= 1 – $$\frac{1}{2}$$ (cos 2α + cos 2β) – 1 + cos2 γ
= $$\frac{-1}{2}$$ × 2 cos(α + β) cos(α – β) + cos2 γ
= cos γ cos (α – β) + cos2 γ …..[∵ α + β + γ = π]
= cos γ [cos (α – β) + cos γ]
= cos γ [cos (α – β) – cos (α + β)]
= 2 sin α sin β cos γ

Question 8.
Let 0 < A, B < $$\frac{\pi}{2}$$ satisfying the equation 3sin2 A + 2sin2 B = 1 and 3sin 2A – 2sin 2B = 0, then A + 2B is equal to ________
(a) π
(b) $$\frac{\pi}{2}$$
(c) $$\frac{\pi}{4}$$
(d) 2π
(b) $$\frac{\pi}{2}$$
Hint:
3 sin 2A – 2sin 2B = 0
sin 2B = $$\frac{3}{2}$$ sin 2A …….(i)
3 sin2 A + 2 sin2 B = 1
3 sin2 A = 1 – 2 sin2 B
3 sin2 A = cos 2B ……(ii)
cos(A + 2B) = cos A cos 2B – sin A sin 2B
= cos A (3 sin2 A) – sin A ($$\frac{3}{2}$$ sin 2A) …..[From (i) and (ii)]
= 3 cos A sin2 A – $$\frac{3}{2}$$ (sin A) (2 sin A cos A)
= 3 cos A sin2 A – 3 sin2 A cos A
= 0
= cos $$\frac{\pi}{2}$$
∴ A + 2B = $$\frac{\pi}{2}$$ ……..[∵ 0 < A + 2B < $$\frac{3 \pi}{2}$$]

Question 9.
In ∆ABC if cot A cot B cot C > 0, then the triangle is ________
(a) acute-angled
(b) right-angled
(c) obtuse-angled
(d) isosceles right-angled
(a) acute angled
Hint:
cot A cot B cot C > 0
Case I:
cot A, cot B, cot C > 0
∴ cot A > 0, cot B > 0, cot C > 0
∴ 0 < A < $$\frac{\pi}{2}$$, 0 < B < $$\frac{\pi}{2}$$, 0 < C < $$\frac{\pi}{2}$$
∴ ∆ABC is an acute angled triangle.
Case II:
Two of cot A, cot B, cot C < 0
0 < A, B, C < π and two of cot A, cot B, cot C < 0
∴ Two angles A, B, C are in the 2nd quadrant which is not possible.

Question 10.
The numerical value of tan 20° tan 80° cot 50° is equal to ________
(a) √3
(b) $$\frac{1}{\sqrt{3}}$$
(c) 2√3
(d) $$\frac{1}{2 \sqrt{3}}$$
(a) √3
Hint:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)

= tan 3(20°)
= tan 60°
= √3
= R.H.S.

II. Prove the following.

Question 1.
tan 20° tan 80° cot 50° = √3
Solution:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)

= tan 3(20°)
= tan 60°
= √3
= R.H.S.

Question 2.
If sin α sin β – cos α cos β + 1 = 0, then prove that cot α tan β = -1.
Solution:
sin α sin β – cos α cos β + 1 = 0
∴ cos α cos β – sin α sin β = 1
∴ cos (α + β) = 1
∴ α + β = 0 ……[∵ cos 0 = 1]
∴ β = -α
L.H.S. = cot α tan β
= cot α tan(-α)
= -cot α tan α
= -1
= R.H.S.

Question 3.
$$\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=\frac{1}{16}$$
Solution:

Question 4.
$$\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=\frac{1}{8}$$
Solution:

Question 5.
cos 12° + cos 84° + cos 156° + cos 132° = $$-\frac{1}{2}$$
Solution:

Question 6.
$$\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x$$
Solution:

Question 7.
$$\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$$
Solution:

Question 8.
sin2 6x – sin2 4x = sin 2x sin 10x
Solution:

Question 9.
cos2 2x – cos2 6x = sin 4x sin 8x
Solution:

Question 10.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution:

Question 11.
$$\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$$
Solution:

Question 12.
If sin 2A = λ sin 2B, then prove that $$\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda+1}{\lambda-1}$$
Solution:

Question 13.
$$\frac{2 \cos 2 A+1}{2 \cos 2 A-1}$$ = tan (60° + A) tan (60° – A)
Solution:

Question 14.
tan A + tan (60° + A) + tan (120° + A) = 3 tan 3A
Solution:

Question 15.
3 tan6 10° – 27 tan4 10° + 33 tan2 10° = 1
Solution:

Question 16.
cosec 48° + cosec 96° + cosec 192° + cosec 384° = 0
Solution:
L.H.S. = cosec 48° + cosec 96° + cosec 192° + cosec 384°
= cosec 48° + cosec (180° – 84°) + cosec (180° + 12°) + cosec (360° + 24°)
= cosec 48° + cosec 84° + cosec (-12°) + cosec 24°

Question 17.
3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x) = 13
Solution:
(sin x – cos x)4
= [(sin x – cos x)2]2
= (sin2 x + cos2 x – 2 sin x cos x)2
= (1 – 2 sin x cosx)2
= 1 – 4 sin x cos x + 4 sin2 x cos2 x
(sin x + cos x)2 = sin2 x + cos2 x + 2 sin x cos x = 1 + 2 sin x cos x
sin6 x + cos6 x
= (sin2 x)3 + (cos2 x)3
= (sin2 x + cos2 x)3 – 3 sin2 x cos2 x (sin2 x + cos2 x) …..[∵ a3 + b3 = (a + b)3 – 3ab(a + b)]
= 13 – 3 sin2 x cos2 x (1)
= 1 – 3 sin2 x cos2 x
L.H.S. = 3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x)
= 3(1 – 4 sin x cos x + 4 sin2 x cos2 x) + 6(1 + 2 sin x cos x) + 4(1 – 3 sin2 x cos2 x)
= 3 – 12 sin x cos x + 12 sin2 x cos2 x + 6 + 12 sin x cos x + 4 – 12 sin2 x cos2 x
= 13
= R.H.S.

Question 18.
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
Solution:
We have to prove that,
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
i.e., to prove,
cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A = 0

∴ cot θ – tan θ = 2 cot 2θ …..(i)
L.H.S. = cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A
= 2 cot 2A – 2 tan 2A – 4 tan 4A – 8 cot 8A …..[From (i)]
= 2(cot 2A – tan 2A) – 4 tan 4A – 8 cot 8A
= 2 × 2 cot 2(2A) – 4 tan 4A – 8 cot 8A ……[From (i)]
= 4(cot 4A – tan 4A) – 8 cot 8A
= 4 × 2 cot 2(4A) – 8 cot 8A ……[From (i)]
= 8 cot 8A – 8 cot 8A = 0
= R.H.S.
Alternate Method:

Question 19.
If A + B + C = $$\frac{3 \pi}{2}$$, then cos 2A + cos 2B + cos 2C = 1 – 4 sin A sin B sin C
Solution:

Question 20.
In any triangle ABC, sin A – cos B = cos C. Show that ∠B = $$\frac{\pi}{2}$$.
Solution:
sin A – cos B = cos C
∴ sin A = cos B + cos C

A = B – C ………(i)
In ∆ABC,
A + B + C = π
∴ B – C + B + C = π
∴ 2B = π
∴ B = $$\frac{\pi}{2}$$

Question 21.
$$\frac{\tan ^{3} x}{1+\tan ^{2} x}+\frac{\cot ^{3} x}{1+\cot ^{2} x}$$ = sec x cosec x – 2 sin x cos x
Solution:

Question 22.
sin 20° sin 40° sin 80° = $$\frac{\sqrt{3}}{8}$$
Solution:
L.H.S. = sin 20°. sin 40°. sin 80°
= sin 20°. sin 40°. sin 80°
= $$\frac{1}{2}$$ (2 . sin 40°. sin 20°) . sin 80°
= $$\frac{1}{2}$$ [cos(40° – 20°) – cos (40° + 20°)] . sin 80°
= $$\frac{1}{2}$$ (cos 20° – cos 60°) sin 80°
= $$\frac{1}{2}$$ . cos 20° . sin 80° – $$\frac{1}{2}$$ . cos 60° . sin 80°
= $$\frac{1}{2 \times 2}$$ (2 sin 80° . cos 20°) – $$\frac{1}{2 \times 2}$$ . sin 80°
= $$\frac{1}{4}$$ [sin(80° + 20°) + sin (80° – 20°)] – $$\frac{1}{2}$$ . sin 80°

Question 23.
sin 18° = $$\frac{\sqrt{5}-1}{4}$$
Solution:
Let θ = 18°
∴ 5θ = 90°
∴ 2θ + 3θ = 90°
∴ 2θ = 90° – 3θ
∴ sin 2θ = sin (90° – 3θ)
∴ sin 2θ = cos 3θ
∴ 2 sin θ cos θ = 4 cos3 θ – 3 cos θ
∴ 2 sin θ = 4 cos2 θ – 3 …..[∵ cos θ ≠ 0]
∴ 2 sin θ = 4 (1 – sin2 θ) – 3
∴ 2 sin θ = 1 – 4 sin2 θ
∴ 4 sin2 θ + 2 sin θ – 1 = 0
∴ sin θ = $$\frac{-2 \pm \sqrt{4+16}}{8}$$
= $$\frac{-2 \pm 2 \sqrt{5}}{8}$$
= $$\frac{-1 \pm \sqrt{5}}{4}$$
Since, sin 18° > 0
∴ sin 18°= $$\frac{\sqrt{5}-1}{4}$$

Question 24.
cos 36° = $$\frac{\sqrt{5}+1}{4}$$
Solution:
We know that,
cos 2θ = 1 – 2 sin2 θ
cos 36° = cos 2(18°)
= 1 – 2 sin2 18°

∴ cos 36° = $$\frac{\sqrt{5}+1}{4}$$

Question 25.
sin 36° = $$\frac{\sqrt{10-2 \sqrt{5}}}{4}$$
Solution:
We know that, sin2 θ = 1 – cos2 θ
sin2 36° = 1 – cos2 36°
= 1 – $$\left(\frac{\sqrt{5}+1}{4}\right)^{2}$$
= $$\frac{16-(5+1+2 \sqrt{5})}{16}$$
= $$\frac{10-2 \sqrt{5}}{16}$$
∴ sin 36° = $$\frac{\sqrt{10-2 \sqrt{5}}}{4}$$ ……[∵ sin 36° is positive]

Question 26.
$$\sin \frac{\pi^{c}}{8}=\frac{1}{2} \sqrt{2-\sqrt{2}}$$
Solution:

Question 27.
tan $$\frac{\pi}{8}$$ = √2 – 1
Solution:

Question 28.
tan 6° tan 42° tan 66° tan 78° = 1
Solution:

Question 29.
sin 47° + sin 61° – sin 11° – sin 25° = cos 7°
Solution:

Question 30.
√3 cosec 20° – sec 20° = 4
Solution:

Question 31.
In ∆ABC, ∠C = $$\frac{2 \pi}{3}$$, then prove that cos2 A + cos2 B – cos A cos B = $$\frac{3}{4}$$.
Solution: