Trigonometric Functions MCQ AP Inter 1st Year Maths Chapter 3

Referring to the AP Inter 1st Year Maths Study Material Chapter Trigonometric Functions MCQ Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Trigonometric Functions MCQ

Multiple Choice Questions

Question 1.
The value of cos 5π is
1) 0
2) 1
3) – 1
4) None of these
Solution:
3) – 1

Explanation:
Cos 5π = cos (2 × 2π + π)
= cos π = – 1.

Question 2.
The value of cos 1° cos 2° cos 3° ………………………. cos 179° is
1) \(\frac{1}{\sqrt{2}}\)
2) 0
3) 1
4) – 1
Solution:
2) 0

Explanation:
cos 1° cos 2° cos 3° ………………………. cos 179°
= cos 1° . cos 2° ….. cos 3° ………………. cos 90° ………… cos 179°
= cos 1° . cos 2° …………. 0 …………. cos 179° = 0

Question 3.
If sin θ + cosec θ = 2, then sin² θ + cosec² θ is equal to
1) 1
2) 4
3) 2
4) None of these
Solution:
3) 2

Explanation:
Given sin θ + cosec θ = 2
⇒ sin θ + \(\frac{1}{\sin \theta}\) = 2
⇒ sin² θ + 1 = 2 sin θ
⇒ sin² θ – 2 sin θ + 1 = 0
⇒ sin θ – 1 = 0
⇒ sin θ = 1
Now sin² θ + cosec² θ = (1)² + (1)²
= 1 + 1 = 2

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

Explanation:
tan (θ + ϕ) = \(\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}\)
= \(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \cdot \frac{1}{3}}\)
= \(\frac{3+2}{6-1}\)
= 1 [∴ (θ + ϕ) = \(\frac{\pi}{4}\)]

Question 5.
The value of \(\frac{1-\tan ^2 15}{1+\tan ^2 15}\) is
1) 1
2) √3
3) \(\frac{\sqrt{3}}{2}\)
4) 2
Solution:
3) \(\frac{\sqrt{3}}{2}\)

Explanation:
\(\frac{1-\tan ^2 15}{1+\tan ^2 15}\) = cos (2 × 15°)
= cos 30°
= \(\frac{\sqrt{3}}{2}\)

Question 6.
The value of sin (45° + θ) – cos (45° – θ) is
1) 2 cos θ
2) 2 sin θ
3) 1
4) 0
Solution:
4) 0

Explanation:
sin (45° + θ) – cos (45° – θ) = (sin 45° cos θ + cos 45° sin θ) – (cos 45° cos θ + sin 45° sin θ)
= [\(\frac{1}{\sqrt{2}}\) cos θ + \(\frac{1}{\sqrt{2}}\) sin θ] – [\(\frac{1}{\sqrt{2}}\) cos θ + \(\frac{1}{\sqrt{2}}\) sin θ] = 0

Question 7.
The value of cot (\(\frac{\pi}{4}\) + θ) cot (\(\frac{\pi}{4}\) – θ) is
1) – 1
2) 0
3) 1
4) None of these
Solution:
3) 1

Explanation:
cot (\(\frac{\pi}{4}\) + θ) cot (\(\frac{\pi}{4}\) – θ)

Question 8.
cos 2θ cos 2ϕ + sin² (θ – ϕ) – sin² (θ + ϕ) is equal to
1) sin 2 (θ + ϕ)
2) cos 2 (θ + ϕ)
3) sin 2 (θ – ϕ)
4) cos 2 (θ – ϕ)
Solution:
2) cos 2 (θ + ϕ)

Explanation:
cos 2θ cos 2ϕ + sin² (θ – ϕ) – sin² (θ + ϕ)
= cos 2θ cos 2ϕ + sin [(θ – ϕ) + (θ + ϕ)] sin [(θ – ϕ) – (θ + ϕ)]
= cos 2θ cos 2ϕ + sin 2θ sin (- 2ϕ)
= cos 2θ cos 2ϕ – sin 2θ sin 2ϕ
= cos (2θ + 2ϕ)
= cos 2 (θ + ϕ)

Question 9.
The value of sin 50° – sin 70° + sin 10° is equal to
1) 1
2) 0
3) ½
4) 2
Solution:
2) 0

Explanation:
sin 50° – sin 70° + sin 10° = 2 cos \(\left[\frac{50^{\circ}+70^{\circ}}{2}\right]\) sin \(\left[\frac{50^{\circ}-70^{\circ}}{2}\right]\) + sin 10°
= 2 cos 60° sin (- 10°) + sin 10°
= – sin 10° + sin 10° = 0.

Question 10.
If sin θ + cos θ = 1 then the value of sin 2θ is equal to
1) 1
2) ½
3) 0
4) – 1
Solution:
3) 0

Explanation:
sin θ + cos θ = 1
⇒ (sin θ + cos θ)² = 1
⇒ sin² θ + cos² θ + 2 sin θ cos θ = 1
⇒ 1 + sin 2θ = 1
⇒ sin 2θ = 0.