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

Answer:

(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) \(\frac{x y}{x-y}\)

Answer:

(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

Answer:

(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) \(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\)

Answer:

(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

Answer:

(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) \(\frac{1}{256}\)

Answer:

(b) \(\frac{1}{64}\)

Hint:

Question 7.

If α + β + γ = π, then the value of sin^{2} α + sin^{2} β – sin^{2} 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

Answer:

(c) 2 sin α sin β cos γ

Hint:

sin^{2} α + sin^{2} β – sin^{2} 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 + cos^{2} γ

= \(\frac{-1}{2}\) × 2 cos(α + β) cos(α – β) + cos^{2} γ

= cos γ cos (α – β) + cos^{2} 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 3sin^{2} A + 2sin^{2} 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

Answer:

(b) \(\frac{\pi}{2}\)

Hint:

3 sin 2A – 2sin 2B = 0

sin 2B = \(\frac{3}{2}\) sin 2A …….(i)

3 sin^{2} A + 2 sin^{2} B = 1

3 sin^{2} A = 1 – 2 sin^{2} B

3 sin^{2} A = cos 2B ……(ii)

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

= cos A (3 sin^{2} A) – sin A (\(\frac{3}{2}\) sin 2A) …..[From (i) and (ii)]

= 3 cos A sin^{2} A – \(\frac{3}{2}\) (sin A) (2 sin A cos A)

= 3 cos A sin^{2} A – 3 sin^{2} 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

Answer:

(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) \(\frac{1}{2 \sqrt{3}}\)

Answer:

(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.

sin^{2} 6x – sin^{2} 4x = sin 2x sin 10x

Solution:

Question 9.

cos^{2} 2x – cos^{2} 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 tan^{6} 10° – 27 tan^{4} 10° + 33 tan^{2} 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(sin^{6} x + cos^{6} x) = 13

Solution:

(sin x – cos x)^{4}

= [(sin x – cos x)^{2}]^{2}

= (sin^{2} x + cos^{2} x – 2 sin x cos x)^{2}

= (1 – 2 sin x cosx)^{2}

= 1 – 4 sin x cos x + 4 sin^{2} x cos^{2} x

(sin x + cos x)^{2} = sin^{2} x + cos^{2} x + 2 sin x cos x = 1 + 2 sin x cos x

sin^{6} x + cos^{6} x

= (sin^{2} x)^{3} + (cos^{2} x)^{3}

= (sin^{2} x + cos^{2} x)^{3} – 3 sin^{2} x cos^{2} x (sin^{2} x + cos^{2} x) …..[∵ a^{3} + b^{3} = (a + b)^{3} – 3ab(a + b)]

= 1^{3} – 3 sin^{2} x cos^{2} x (1)

= 1 – 3 sin^{2} x cos^{2} x

L.H.S. = 3(sin x – cos x)^{4} + 6(sin x + cos x)^{2} + 4(sin^{6} x + cos^{6} x)

= 3(1 – 4 sin x cos x + 4 sin^{2} x cos^{2} x) + 6(1 + 2 sin x cos x) + 4(1 – 3 sin^{2} x cos^{2} x)

= 3 – 12 sin x cos x + 12 sin^{2} x cos^{2} x + 6 + 12 sin x cos x + 4 – 12 sin^{2} x cos^{2} 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 cos^{3} θ – 3 cos θ

∴ 2 sin θ = 4 cos^{2} θ – 3 …..[∵ cos θ ≠ 0]

∴ 2 sin θ = 4 (1 – sin^{2} θ) – 3

∴ 2 sin θ = 1 – 4 sin^{2} θ

∴ 4 sin^{2} 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 sin^{2} θ

cos 36° = cos 2(18°)

= 1 – 2 sin^{2} 18°

∴ cos 36° = \(\frac{\sqrt{5}+1}{4}\)

Question 25.

sin 36° = \(\frac{\sqrt{10-2 \sqrt{5}}}{4}\)

Solution:

We know that, sin^{2} θ = 1 – cos^{2} θ

sin^{2} 36° = 1 – cos^{2} 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 cos^{2} A + cos^{2} B – cos A cos B = \(\frac{3}{4}\).

Solution: