ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

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ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Choose the correct answer from the given four options in questions (1 to 44) :

Question 1.
1 radian is approximately equal to
(a) 57°16’
(b) 4718’30”
(c) 53°17’45”
(d) 43°16’
Solution:
(a) 57°16’

We know that,
π radians = 180°

Question 2.
1° is approximately equal to
(a) 0.001746 radians
(b) 0.01746 radians
(c) 0.0001746 radians
(d) 0.1746 radians
Solution:
(b) 0.01746 radians

We know that,
π radians = 180°
⇒ 1° = \(\frac{\pi}{180}\)
= \(\frac{22}{7 \times 180}\)
⇒ 1° = 0.0 174603 radians.

Question 3.
If θ lies in second quadrant, then the quadrant in which lies is
(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant
Solution:
(d) IV quadrant

Given θ lies in 2nd quadrant
∴ 90° ≤ θ ≤ 180°
⇒ – 45° ≥ – \(\frac{\theta}{2}\) ≥ 90°
⇒ – 90° ≤ – \(\frac{\theta}{2}\) ≤ 45°
∴ θ lies in IV quadrant.

Question 4.
The angle in degree measure between two hands of a clock at 8 : 30 p.m. is
(a) 55°
(b) 66°
(c) 75°
(d) 80°
Solution:
(c) 75°

angle trace out by hour hand in 12 hours = 360°
angle trace out by hour hand in 8\(\frac{1}{2}\) hours = \(\frac{360^{\circ}}{12} \times \frac{17}{2}\)
Φ = 255°
Also, angle trace out by minute hand in 60 minutes = 300°
∴ angle trace out by minute hand in 30 minutes = θ
= \(\frac{360^{\circ}}{60}\) × 30 = 180°
Thus required angle = Φ – θ
= 255° – 180° = 75°

Question 5.
The angle subtended by an arc of length 20 cm at the centre of circle when radius is 14 cm is
(a) \(\frac{5}{7}\) radians
(b) \(\frac{10}{7}\) radians
(c) \(\frac{5}{14}\) radians
(d) \(\frac{7}{10}\) radians
Solution:
(b) \(\frac{10}{7}\) radians

arc length = l = 20 cm ;
r = 14 cm
Since, θ = \(\frac{l}{r}\)
= \(\frac{20}{14}=\frac{10}{7}\) radians

Question 6.
A wheel makes 450 revolutions per hour.
The number of radians through which it turns in one second is
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{6}\)
(d) \(\frac{\pi}{3}\)
Solution:
(a) \(\frac{\pi}{4}\)

Given revolutions made by wheel per hour = 450
∴ No. of revolutions made by wheel in one second = \(\frac{450}{60 \times 60}=\frac{1}{8}\)
Further, distance covered by wheel in one revolution = 2π radians
∴ distance covered by wheel in revolution = \(\frac{\pi}{4}\) radians

Question 7.
If sin x = \(\frac{3}{5}\), then cosx is
(a) \(\frac{4}{5}\) but not \(-\frac{4}{5}\)
(b) \(\frac{4}{5}\) or \(-\frac{4}{5}\)
(c) \(-\frac{4}{5}\) but not \(\frac{4}{5}\)
(d) none of these
Solution:
(b) \(\frac{4}{5}\) or \(-\frac{4}{5}\)

Given sin x = (b) \(\frac{3}{5}\)
∴ cos x = ± \(\sqrt{1-\sin ^2 x}\)
= ± \(\sqrt{1-\frac{9}{25}}= \pm \frac{4}{5}\)

Question 8.
The value of cosec (- 750°) is
(a) \(\frac{2}{\sqrt{3}}\)
(b) – 2
(c) 2
(d) – \(\frac{2}{\sqrt{3}}\)
Solution:
(c) 2

cosec (- 750°) = – cosec 750°
[∵ cosec (- θ) = – cosec θ]
= – cosec (720° + 30°)
= – cosec 30°
= – 2

Question 9.
The value of tan (- \(\frac{15 \pi}{4}\)) is
(a) – 1
(b) 1
(c) \(\frac{1}{\sqrt{3}}\)
(d) \(-\frac{1}{\sqrt{3}}\)
Solution:
(b) 1

tan (- \(\frac{15 \pi}{4}\)) = – tan \(\frac{15 \pi}{4}\)
[∵ tan (- θ) = – tan θ]
= – tan (2π + \(\frac{7 \pi}{4}\))
= – tan \(\frac{7 \pi}{4}\)
= – tan (2π – \(\frac{\pi}{4}\))
= – {- tan \(\frac{\pi}{4}\)}
= tan \(\frac{\pi}{4}\) = 1

Question 10.
The range of 4 + 5 cos x is
(a) [- 1, 9]
(b) (- 1, 9]
(c) (- 1, 9)
(d) [- 1, 9)
Solution:
(a) [- 1, 9]

Since – 1 ≤ cos x ≤ 1
⇒ – 5 ≤ 5 cos x ≤ 5
⇒ 4 – 5 ≤ 4 + 5 cos x ≤ 4 + 5
⇒ – 1 ≤ f (x) ≤ 9
∴ R_f = [- 1, 9]

Question 11.
The domain of 2 sin x cos x is
(a) R – πn
(b) R – (2n + 1) \(\frac{\pi}{2}\)
(c) R
(d) R – \(\frac{n \pi}{2}\)
Solution:
(c) R

Let f (x) = 2 sin x cos x = sin 2x, which is defined ∀ x ∈ R
∴ D_f = R

Question 12.
If x lies in III quadrant and tan x = \(\frac{5}{12}\), then sin x and cos x respectively are
(a) \(\frac{5}{13}, \frac{12}{13}\)
(b) \(-\frac{5}{13}, \frac{12}{13}\)
(c) \(\frac{5}{13},-\frac{12}{13}\)
(d) \(-\frac{5}{13},-\frac{12}{13}\)
Solution:
(d) \(-\frac{5}{13},-\frac{12}{13}\)

Given tan x = \(\frac{5}{12}\)
Since x lies in IIIrd quadrant
∴ sin x, cos x < 0
∴ sec x = – \(\sqrt{1+\tan ^2 x}\)
= – \(\sqrt{1+\left(\frac{5}{12}\right)^2}\)
= \(-\sqrt{1+\frac{25}{144}}=-\frac{13}{12}\)
cos x = – \(\frac{12}{13}\)
∴ sin x = tan x cos x
= \(\frac{5}{12} \times\left(-\frac{12}{13}\right)\)
= – \(\frac{5}{13}\)

Question 13.
The value of 2 sin² \(\frac{\pi}{6}\) + cosec \(\frac{7 \pi}{6}\) . cos² \(\frac{\pi}{32}\) is equal to
(a) 1
(b) \(\frac{3}{2}\)
(c) – 1
(d) 2
Solution:
(b) \(\frac{3}{2}\)

2 sin² \(\frac{\pi}{6}\) + cosec \(\frac{7 \pi}{6}\) . cos² \(\frac{\pi}{32}\) = \(2\left(\sin \frac{\pi}{6}\right)^2+\left\{\ {cosec}\left(\pi+\frac{\pi}{6}\right)\right\}^2\left(\cos \frac{\pi}{3}\right)^2\)
= \(2\left(\frac{1}{2}\right)^2+\left\{-\ {cosec} \frac{\pi}{6}\right\}^2\left(\frac{1}{2}\right)^2\)
= \(\frac{1}{2}+(-2)^2 \times \frac{1}{4}=\frac{3}{2}\)

Question 14.
The value of \(\cos \left(\frac{3 \pi}{2}+x\right) \cdot \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]\) is
(a) – 1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1

cos (\(\frac{3 \pi}{2}\) + x) . cos (2π + x) [cot (\(\frac{3 \pi}{2}\) – x) + cot (2π + x)]
= sin x . cos x [tan x + cot x]
= sin x cos x \(\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right]\)
= \(\frac{\sin x \cos x\left(\sin ^2 x+\cos ^2 x\right)}{\sin x \cos x}\) = 1

Question 15.
If sin θ + cosec θ = 2, then sin2 θ + cosec2 θ is equal to
(a) 1
(b) 4
(c) 2
(d) 6
Solution:
(c) 2

Given sin θ + cosec θ = 2 …………………….( 1)
∴ sin² θ + cosec² θ = (sin θ + cosec θ)² – 2 sin θ cosec θ
= 2² – 2 = 2 [using eqn. (1)]

Question 16.
If f (x) = cos² x + sec² x, then
(a) f (x)< 1
(b) f(x) = 1
(c) 2< f (x) < 1
(d) f (x) ≥ 2
Solution:
(d) f (x) ≥ 2

f(x) = cos² x + sec² x
= (cos x – sec x)² + 2 cos x sec x
= (cos x – sec x)² + 2 ≥ 2
[∵ (cos x – sec x)² ≥ 0 ∀ x ∈ R]

Question 17.
If tan θ = \(\frac{1}{2}\) and tan = \(\frac{1}{3}\), then the value of (θ + Φ) is
(a) \(\frac{\pi}{6}\)
(b) π
(c) 0
(d) \(\frac{\pi}{4}\)
Solution:
(d) \(\frac{\pi}{4}\)

tan (θ + Φ) = \(\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}\)
= \(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\)
= \(\frac{\frac{5}{6}}{1-\frac{1}{6}}\)
= \(\frac{\frac{5}{6}}{\frac{5}{6}}\)
= 1
θ + Φ = \(\frac{\pi}{4}\)

Question 18.
Which of the following is not correct?
(a) sin θ = – \(\frac{1}{5}\)
(b) cos θ = 1
(c) sec θ = \(\frac{1}{2}\)
(d) tan θ = 20
Solution:
(c) sec θ = \(\frac{1}{2}\)

since |sec θ| ≥ 1
∴ sec θ = \(\frac{1}{2}\) is not possible.

Question 19.
The value of cos 1°. cos 2°. cos 3°………………. cos 179° is
(a) \(\frac{1}{\sqrt{2}}\)
(b) 0
(c) 1
(d) – 1
Solution:
(b) 0

cos 1°. cos 2°………………. cos 90°………………. cos 179° = 0
[∵ cos 90° = 0]

Question 20.
The value of tan 10. tan 2°. tan 3° ……………… tan 89° is
(a) 0
(b) 1
(c) \(\frac{1}{2}\)
(d) 2
Solution:
(b) 1

(tan 1° tan 2° tan 3° ………………. tan 44°) tan 45° (tan 46° tan 47° …………… tan 89°)
= (tan 1° tan 2° …. tan 44°) × 1 × (tan (90° – 44°) tan (90° – 43°) …. tan (90° – 1°)]
= (tan 1° tan 2° ………………. tan 44°) (cot 44° cot 43° ……………….. cot 1°)
= (tan 1° cot 1°) (tan 2° cot 2°) ……………. (tan 44° cot 44°) –
= 1 × 1 × ……………. × 1 = 1
[∵ tan θ . cot θ = 1]

Question 21.
The value of tan 75° – cot 75° is equal to
(a) 2√3
(b) 2 + √3
(c) 2 – √3
(d) 1
Solution:

tan 75° – cot 75° = tan 75°
= tan 75° – \(\frac{1}{\tan 75^{\circ}}\)

Question 22.
Which of the following is correct?
(a) sin 1° > sin 1
(b) sin 1° < sin 1
(c) sin 1° = sin 1
(d) sin 1° = \(\frac{\pi}{180}\) sin 1
Solution:
(b) sin 1° < sin 1

Since 1 radian = 57° 16’
Thus 1° < 1 radian
⇒ sin 1° < sin 1
[∵ since sin θ be an increasing function in first quadrant]

Question 23.
The value of tan 5A – tan 3A – tan 2A is equal to
(a) tan 5A . tan 3A. tan 2A
(b) – tan 5A . tan3A . tan 2A
(c) tan 3A. tan 2A – tan 2A. tan 5A – tan 5A tan 2A
(d) none of these
Solution:
(a) tan 5A . tan 3A. tan 2A

tan 5A = tan (3A + 2A)
= \(\frac{\tan 3 \mathrm{~A}+\tan 2 \mathrm{~A}}{1-\tan 3 \mathrm{~A} \cdot \tan 2 \mathrm{~A}}\)
= tan 5A (1 – tan 3A tan 2A)
= tan 3A + tan 2A
= tan 5A – tan 5A – tan 2A
= tan 5A tan 3A tan 2A

Question 24.
The value of sin (45° + θ) – cos (45° – θ) is
(a) 2 cos θ
(b) 2 sin θ
(c) 1
(d) 0
Solution:
(d) 0

sin (45° + θ) – cos (45° – θ) = sin 45° cos θ + cos 45° sin θ – [cos 45° cos θ + sin 45° sin θ]
[∵ sin (A + B) = sin A cos B + cos A sin B
and cos (A – B) = cos A cos B + sin A sin B]
= \(\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta-\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta\)
= 0

Question 25.
The value of cot (\(\frac{\pi}{4}\) + θ) . cot (\(\frac{\pi}{4}\) – θ) is
(a) – 1
(b) 0
(c) 1
(d) not defined
Solution:
(c) 1

Question 26.
cos 2θ . cos 2Φ + sin² (θ – Φ) – sin² (θ + Φ) is equal to
(a) sin 2 (θ + Φ)
(b) cos 2 (θ + Φ)
(c) sin 2 (θ – Φ)
(d) cos 2 (- Φ)
Solution:
(b) cos 2 (θ + Φ)

cos 2θ cos 2Φ + sin² (θ – Φ) – sin² (θ + Φ) = cos 2θ cos 2Φ + sin (θ – Φ + θ + Φ) sin (θ – Φ – θ – Φ)
[∵ sin² A – sin² B = sin (A + B) sin (A – B)]
= cos 2θ cos 2Φ + sin 2θ sin (- 2Φ)
= cos 2θ cos 2Φ – sin 2θ sin 2Φ
= cos (2θ + 2Φ)

Question 27.
If for real x, cos θ = x + \(\frac{1}{x}\), then
(a) θ is an acute angle
(b) is right angle
(c) θ is an obtuse angle
(d) no value of θ is possible
Solution:
(d) no value of θ is possible

Given cos θ = x + \(\frac{1}{x}\)
⇒ x² – x cos θ + 1 = 0
Here D = cos² θ – 4 < 0
[∵ 0 ≤ cos² θ ≤ 1]
∴ x is non-real, which contradicts the given fact that x is real.
Thus there is no value of θ for which the given equation holds.

Question 28.
If tan A = \(\frac{1}{2}\), tan B = \(\frac{1}{3}\) then tan (2A + B) is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 29.
If α + β = \(\frac{\pi}{3}\), then (1 + tan α) (1 + tan β) is
(a) 1
(b) 2
(c) – 1
(d) – 2
Solution:
(b) 2

Question 30.
If sin θ = – \(\frac{4}{5}\) and θ lies in third quadrant, then the value of cos \(\frac{\theta}{2}\) is
(a) \(\frac{1}{5}\)
(b) \(-\frac{1}{\sqrt{10}}\)
(c) \(-\frac{1}{\sqrt{5}}\)
(d) \(\frac{1}{\sqrt{10}}\)
Solution:
(c) \(-\frac{1}{\sqrt{5}}\)

Since 180° ≤ θ ≤ 270°
⇒ 9o° ≤ \(\frac{\theta}{2}\) ≤ 135°
∴ \(\frac{\theta}{2}\) lies in 2nd quadrant
∴ cos \(\frac{\theta}{2}\) < 0
given sin θ = – \(\frac{4}{5}\) and θ lies in third quadrant
∴ cos θ < 0

Question 31.
The greatest value of sin x . cos x ¡s
(a) 1
(b) 2
(c) √2
(d) \(\frac{1}{2}\)
Solution:
(d) \(\frac{1}{2}\)

sin x cos x = \(\frac{1}{2}\) (2 sin x cos x)
= \(\frac{1}{2}\) sin 2x
= f (x)
since – 1 ≤ sin 2x ≤ 1
⇒ – \(\frac{1}{2}\) ≤ \(\frac{1}{2}\) sin 2x ≤ \(\frac{1}{2}\)
– \(\frac{1}{2}\) ≤ f (x) ≤ \(\frac{1}{2}\)
∴ greatest value of f (x) = \(\frac{1}{2}\)

Question 32.
a cos x + b sin x lies between
(a) a and b
(b) – (a² + b²) and (a² + b²)
(c) \(-\sqrt{a^2+b^2} \text { and } \sqrt{a^2+b^2}\)
(d) \(-\sqrt{a+b} \text { and } \sqrt{a+b}\)
Solution:
(c) \(-\sqrt{a^2+b^2} \text { and } \sqrt{a^2+b^2}\)

put a = r cos α
and b = r sin α
On squaring and adding eqn. (1) and (2) ; we have
a² + b² = r²
⇒ r = \(\sqrt{a^2+b^2}\) (∵ r > 0)
On dividing eqn. (2) by eqn. (1) ; we have
tan α = \(\frac{b}{a}\)
∴ a cos x + b sin x = r cos α cos x + r sin x sin α
= r cos (x – α)
since – 1 ≤ cos (x – α) ≤ 1
⇒ – r ≤ r cos (x – α) ≤ r
⇒ – \(\sqrt{a^2+b^2}\) ≤ a cos x + b sin x ≤ \(\sqrt{a^2+b^2}\).

Question 33.
Number of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2π] is
(a) 0
(b) 1
(c) 2
(d) 3
Solution
(c) 2

Given eqn. be, tan x + sec x 2 cos x
⇒ \(\frac{\sin x+1}{\cos x}\) = 2 cos x
⇒ sin x + 1 = 2 cos² x, cosx ≠ 0
⇒ 1 + sin x = 2 (1 – sin² x), x ≠ \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)
2 sin² x + sin x – 1 = 0
⇒ sin x = \(\frac{-1 \pm \sqrt{1+8}}{4}\)
= \(\frac{-1 \pm 3}{4}\)
⇒ sin x = \(\frac{1}{2}\), – 1 ; x ≠ \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)
∴ sin x = \(\frac{1}{2}\) = \(\frac{\pi}{6}\), sin (π – \(\frac{\pi}{6}\))
[∵ x ∈ [0, 2π])
and sin x = – 1
= sin \(\frac{3 \pi}{2}\), sin (- \(\frac{\pi}{2}\)) but x ∈ [0, 2π]
⇒ x = \(\frac{3 \pi}{2}\), – \(\frac{3 \pi}{2}\) which are not possible.

Question 34.
If cos x = – \(\frac{3}{5}\) and π < x < \(\frac{3 \pi}{2}\), then \(\) is equal to (a) \(\frac{1}{6}\) (b) – \(\frac{1}{3}\) (c) – \(\frac{1}{6}\) (d) \(\frac{2}{3}\) Solution: Given cos x = – \(\frac{3}{5}\) and x lies in 3rd quadrant. ∴ tan x, cot x > 0 and cosec x < 0
∴ sec x = \(\frac{1}{\cos x}=-\frac{5}{3}\)
and tan x = \(\sqrt{\sec ^2 x-1}\)
= \(\sqrt{\frac{25}{9}-1}=\frac{4}{3}\)
and cot x = \(\frac{3}{4}\) ;
sin x = tan x . cos x
= \(\frac{4}{3} \times\left(-\frac{3}{5}\right)=-\frac{4}{5}\)
∴ cosec x = – \(\frac{5}{4}\)
Thus, \(\frac{\ {cosec} x+\cot x}{\sec x-\tan x}=\frac{-\frac{5}{4}+\frac{3}{4}}{-\frac{5}{3}-\frac{4}{3}}\)
= \(\frac{-\frac{2}{4}}{-\frac{9}{3}}=\frac{1}{6}\)

Question 35.
The value of sin (π + x). sin (π – x) . coscc² x is
(a) 0
(b) 1
(c) – \(\frac{1}{6}\)
(d) \(\frac{2}{3}\)
Solution:
(c) – \(\frac{1}{6}\)

sin (π + x) sin(π – x) . cosec² x = – sin x sin x cosec² x = – 1

Question 36.
The value of \(3 \sin \frac{\pi}{6} \cdot \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cdot \cot \frac{\pi}{4}\) is
(a) – 1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1

\(3 \sin \frac{\pi}{6} \cdot \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cdot \cot \frac{\pi}{4}\)
= \(3 \times \frac{1}{2} \times \frac{2}{1}-4 \sin \left(\pi-\frac{\pi}{6}\right) \times 1\)
= 3 – 4 sin \(\frac{\pi}{6}\)
= 3 – 4 sin × \(\frac{1}{2}\)
= 3 – 2 = 1

Question 37.
If sin α = k sin β, then tan \(\left(\frac{\alpha-\beta}{2}\right)\) . cot \(\left(\frac{\alpha+\beta}{2}\right)\) is equal to
(a) \(\frac{k-1}{k+1}\)
(b) \(\frac{k+1}{k-1}\)
(c) \(\frac{1+k}{1-k}\)
(d) \(\frac{1-k}{1+k}\)
Solution:
(a) \(\frac{k-1}{k+1}\)

Question 38.
If sin x = \(\frac{1}{3}\), then the value of sin 3x is
(a) 1
(b) 0
(c) \(\frac{23}{27}\)
(d) – \(\frac{23}{27}\)
Solution:

Given sin x = \(\frac{1}{3}\)
Thus, sin 3x = 3 sin x – 4 sin³ x
= \(3 \times \frac{1}{3}-4\left(\frac{1}{3}\right)^3\)
= \(1-\frac{4}{27}=\frac{23}{27}\)

Question 39.
The value of cos² 48° – sin² 12° is
(a) \(\frac{\sqrt{5}+1}{8}\)
(b) \(\frac{\sqrt{5}-1}{8}\)
(c) \(\frac{\sqrt{5}+1}{4}\)
(d) \(\frac{\sqrt{5}-1}{4}\)
Solution:

cos² 48° – sin² 12° = cos (48° + 12°) cos (48° – 12°)
[∵ cos (A + B) cos (A – B) = cos² A – sin² B]
= cos 60° cos 36°
= \(\frac{1}{2} \times \frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}+1}{8}\)

Question 40.
The value of \(\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}\) is
(a) \(-\frac{1}{16}\)
(b) \(-\frac{1}{8}\)
(c) \(\frac{1}{8}\)
(d) \(\frac{1}{16}\)
Solution:

Question 41.
The general solution of the equation tan (2x + \(\frac{\pi}{12}\)) = 0 is
(a) \(\frac{n \pi}{2}-\frac{\pi}{24}\)
(b) nπ + \(\frac{\pi}{12}\)
(c) nπ – \(\frac{\pi}{12}\)
(d) \(\frac{n \pi}{2}+\frac{\pi}{24}\)
Solution:
(a) \(\frac{n \pi}{2}-\frac{\pi}{24}\)

Given tan(2x + \(\frac{\pi}{12}\)) = 0
⇒ 2x + \(\frac{\pi}{12}\) = nπ
[∵ tan θ = 0
⇒ θ = nπ]
⇒ 2x = nπ – \(\frac{\pi}{12}\)
⇒ x = \(\)

Question 42.
The general solution of the equation sin² x . sec x + √3 tan x = 0 is
(a) (2n + 1) \(\frac{\pi}{2}\)
(b) nπ
(c) nπ + \(\frac{\pi}{6}\)
(d) nπ – \(\frac{\pi}{6}\)
Solution:

Given eqn. be,
sin² x sec x + √3 tan x = 0
⇒ \(\frac{\sin ^2 x}{\cos x}+\frac{\sqrt{3} \sin x}{\cos x}\) = 0
⇒ sin² x + √3 sin x = 0, cos x ≠ 0
⇒ sin x (sin x + √3) = 0, cos x ≠ 0
⇒ sin x = 0 or sin x = – √3, cos x ≠ 0
but |sin x| ≤ 1
∴ sin x = – √3 is not possible.
∴ sin x = 0
⇒ x = nπ and x ≠ nπ + \(\frac{\pi}{2}\)
⇒ x = nπ, n ∈ I

Question 43.
If A lies in the second quadrant and 3 tan A + 4 = 0, then the value of 2 cot A – 5 cos A + sin A is
(a) \(-\frac{53}{10}\)
(b) \(\frac{23}{10}\)
(c) \(\frac{37}{10}\)
(d) \(\frac{7}{10}\)
Solution:
3 tan A + 4 = 0
⇒ tan A = – \(\frac{4}{3}\)
Since A lies in 2nd quadrant
∴ sin A > 0, cos A < 0
sec A = – \(\sqrt{1+\tan ^2 A}\)
= – \(\sqrt{1+\frac{16}{9}}=-\frac{5}{3}\)
⇒ cos A = – \(\frac{3}{5}\)
and sin A = tan A . cos A
= \(-\frac{4}{3} \times\left(-\frac{3}{5}\right)=\frac{4}{5}\)
∴ 2 cot A – 5 cos A + sin A = \(2\left(-\frac{3}{4}\right)-5\left(-\frac{3}{5}\right)+\frac{4}{5}\)
= \(-\frac{3}{2}+3+\frac{4}{5}\)
= \(\frac{-15+30+8}{10}=\frac{23}{10}\)

Question 44.
If tan θ = \(\frac{a}{b}\), then b cos 2θ + a sin 2θ is equal to
(a) a
(b) b
(c) \(\frac{a}{b}\)
(d) \(\frac{b}{a}\)
Solution:
(b) b