# Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.3

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Ex 3.3 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.3

Question 1.
Find the values of:
i. sin $$\frac{\pi}{8}$$
ii. $$\frac{\pi}{8}$$
Solution:
We know that sin2 θ = $$\frac{1-\cos 2 \theta}{2}[/atex] Substituting θ = $\frac{\pi}{8}$$, we get ii. We know that, cos2 θ = $$\frac{1+\cos 2 \theta}{2}$$ Substituting θ = $$\frac{\pi}{8}$$, we get Question 2. Find sin 2x, cos 2x, tan 2x if sec x = $$-\frac{13}{5}$$, $$\frac{\pi}{2}$$ < x < π Solution: sec x = $$-\frac{13}{5}$$, $$\frac{\pi}{2}$$ < x < π We know that Sect2 x = 1 + tan2x tan2x = $$\frac{169}{25}-1=\frac{144}{25}$$ tan x = $$\pm \frac{12}{5}$$ Since $$\frac{\pi}{2}$$ < x < π x lies in the 2nd quadrant. tan x < 0 Question 3. i.  = tan2 θ Solution: L. H. S. = $$\frac{1-\cos 2 \theta}{1+\cos 2 \theta}$$ = $$\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}$$ = 2tan2 θ = R.H.S. ii. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0 Solution: L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cosx)cosx = sin 3x sin x + sin2 x + cos 3x cos x – cos2 x = (cos 3x cos x + sin 3x sin x) — (cos2x — sin2x) = cos (3x – x) – cos 2x = cos 2x – cos 2x = 0 = R.H.S. iii. (cos x + cos y )2 + (sin x + sin y)2 = 4cos2 $$\left(\frac{x-y}{2}\right)$$ Solution: L.H.S. = (cos x + cos y)2 + (sin x + sin y)2 = cos2x + cos2y + 2cos x.cos y + sin2 x + sin2y + 2sin x.siny = (cos2x + sin2x) + (cos2y + sin2y) + 2(cos x.cos y + si x.sin y) = 1 + 1 +2cos(x – y) = 2 + 2 cos (x – y) = 2[1 + cos(x – y)] = 2[2cos2 [($$\left(\frac{x-y}{2}\right)$$)] … [∵ 1 + cos θ = 2 cos2 $$\frac{\theta}{2}$$] = 4 cos2 ($$\frac{x-y}{2}$$) = R.H.S. [ Note: The question has been modified] iv. (cos x – cos y)2 + (sin x – sin y)2 = 4sin2 $$\left(\frac{x-y}{2}\right)$$ Solution: L.H.S. = (cos x – cos y)2 + (sin x – sin y)2 = cos2x + cos2y + 2cos x.cos y + sin2 x + sin2y + 2sin x.siny = (cos2x + sin2x) + (cos2y + sin2y) – 2(cos x.cos y + sin x.sin y) = 1 + 1 – 2cos(x – y) = 2 – 2 cos (x – y) = 2[1 – cos(x – y)] = 2[2sin2 [($$\left(\frac{x-y}{2}\right)$$)] … [∵ 1 – cos θ = 2 sin2 $$\frac{\theta}{2}$$] = 4 sin2 ($$\frac{x-y}{2}$$) = R.H.S. v. tan x + cot x = 2 cosec 2x Solution: L.H.S. = tan x + cot x vi. $$\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}$$ = 2 tan 2x Solution: vii. $$\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 x}}}$$ = 2 cos x Solution: = 2 cos x = R.H.S. [Note : The question has been modified.] viii. 16 sin θ cos θ cos 2θ cos 4θ cos 8θ = sin 16θ Solution: L.H.S. = 16 sin θ cos θ cos 2θ cos 4θ cos 8θ = 8(2sinθ cosθ) cos2θ cos 4θ cos 8θ = 8sin 2θ cos 2θ cos 4θ cos 8θ = 4(2sin 2θ cos 2θ) cos 4θ cos 8θ = 4sin 4θ cos 4θ cos 8θ = 2(2sin 4θ cos 4θ) cos 8θ = 2sin 8θ cos 8θ = sin 16θ = R.H.S. ix. $$= 2 cot 2x Solution: x. [latex]\frac{\cos x}{1+\sin x}=\frac{\cot \left(\frac{x}{2}\right)-1}{\cot \left(\frac{x}{2}\right)+1}$$ Solution: xi. $$\frac{\tan \left(\frac{\theta}{2}\right)+\cot \left(\frac{\theta}{2}\right)}{\cot \left(\frac{\theta}{2}\right)-\tan \left(\frac{\theta}{2}\right)}=\sec \theta$$ Solution: xii. $$\frac{1}{\tan 3 \mathbf{A}-\tan A}-\frac{1}{\cot 3 A-\cot A}$$ = cot 2A Solution: xiii. cos 7° cos 14° cos 28° cos 56° $$\frac{\sin 68^{\circ}}{16 \cos 83^{\circ}}$$ Solution: L.H.S. = cos 7° cos 14° cos 28° cos 56° = $$\frac{1}{2 \sin 7^{\circ}}$$(2sin 7°cos 7°)cos 14°cos 28°cos 56° = $$\frac{1}{2 \sin 7^{\circ}}$$ (sin 14° cos 14° cos 28° cos 56°) …[∵ 2sinθ cosθ = sin 2θ] = [\frac{1}{2\left(2 \sin 7^{\circ}\right)}latex]$ (2sin 14° cos 14°) cos 28° cos 56°
= $$\frac{1}{4 \sin 7^{\circ}}$$(sin 28° cos 28° cos 56°)
= $$\frac{1}{2\left(4 \sin 7^{\circ}\right)}$$(2 sin 28° cos 28°) cos 56°
= $$\frac{1}{8 \sin 7^{\circ}}$$ (sin 56° cos 56°)
= $$\frac{1}{8 \sin 7^{\circ}}$$ (2 sin 56° cos 56°)
= $$\frac{1}{16 \sin 7^{\circ}}$$(sin 112°)
= $$\frac{\sin \left(180^{\circ}-68^{\circ}\right)}{16 \sin \left(90^{\circ}-83^{\circ}\right)}$$
= $$\frac{\sin 68^{\circ}}{16 \cos 83^{\circ}}$$
= R.H.S.

xiv. = $$\frac{\sin ^{2}\left(-160^{\circ}\right)}{\sin ^{2} 70^{\circ}}+\frac{\sin \left(180^{\circ}-\theta\right)}{\sin \theta}$$ = sec2 20°
Solution:

xv. $$\frac{2 \cos 4 x+1}{2 \cos x+1}$$ = (2 cos x – 1)(2 cos 2x – 1)
Solution:

xvi. = cos2 x + cos2 (x + 120°) + cos2(x – 120°) = $$\frac{3}{2}$$
Solution:
L.H.S = cos2 x + cos2 (x + 120°) + cos2(x – 120°) =

$$\frac{3}{2}+\frac{1}{2}$$ [cos 2x + cos(2x + 240°) + cos(2x 240°)]
= $$\frac{3}{2}+\frac{1}{2}$$(cos 2x + cos 2x cos 240°— sin 2x sin 240° + cos 2x cos 240° + sin 2x sin 240°)
= $$\frac{3}{2}+\frac{1}{2}$$(cos 2x + 2 cos 2x cos 240°)
= $$\frac{3}{2}+\frac{1}{2}$$ [cos 2x + 2 cos 2x cos( 180° + 60°)]
= $$\frac{3}{2}+\frac{1}{2}$$ [cos 2x + 2cos 2x(-cos 600)]
= $$\frac{3}{2}+\frac{1}{2}$$ [cos 2x —2 cos 2x($$\frac{1}{2}$$)]
= $$\frac{3}{2}+\frac{1}{2}$$ ( cos 2x – cos 2x)
= $$\frac{3}{2}+\frac{1}{2}$$ (0)
= $$\frac{3}{2}$$ = R.H.S.

xvii. 2 cosec 2x + cosec x = sec cot $$\frac{x}{2}$$
Solution:

xviii. 4 cos x cos ($$\frac{\pi}{3}$$ + x) cos ($$\frac{\pi}{3}$$ – x) = cos 3x

Solution:

= cos3x — 3cos x.sin2x
= cos3 x — 3cos x (1— cos2 x)
= cos3x — 3cos x + 3 cos3x
=4 cos3x — 3cos x
= cos 3x = R.H.S.
INote: The question has been modijìed.I

xix. sin x tan $$\frac{x}{2}$$ + 2cos x = $$\frac{2}{1+\tan ^{2}\left(\frac{x}{2}\right)}$$
Solution:
L.H.S. = sin x tan (x/2)+ 2cos x
= $$\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)$$ + 2cos x
= $$\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)$$ + 2 cos x
= 2sin2 x/2 + 2cosx
= 1 – cosx + 2cosx
= 1 + cos x
=2cos2 x/2
= $$\frac{2}{\sec ^{2} \frac{x}{2}}=\frac{2}{1+\tan ^{2} \frac{x}{2}}$$ =R.H.S.