Multiple choice (Inverse Trigonometry)
choose the correct option in the following question:(class 12 inverse trigonometric functions multiple choice)
Question 1: The value of corresponding to principal branches is
(a) -π/12 (b) 0
(c) π (d) π/3
Answer (c)
Explanation:
Question 2: The value of is
(a) (b)
(c) 1/x (d)
Answer (d)
Explanation:
Let
Since,
Question 3: The value of is
(a) π/9 (b) 5π/9
(c) -5π/9 (d) 7π/18
Answer (d)
Explanation:
Question 4: Let , Then value of θ is
(a) π/3 (b) π/2
(c) 2π/3 (d) – 2π/3
Answer (a)
Explanation:
Question 5: The principal value of is
(a) 2π/5 (b) -2π/5
(c) 3π/5 (d) -3π/5
Answer: (b) -2π/5
Explanation:
{since
}
= –
= –2π/5
Question 6: is equal to:
(a) 1/2 (b) 1/3
(c) -1 (d) 1
Answer: (d) 1
Explanation:
= sin[π/3 – (-π/6)]
= sin(π/3 + π/6)
= sin (π/2)
= 1
Question 7: The domain of is
(a) [0, 1] (b) [– 1, 1]
(c) [-1/2, 1/2] (d) [–2, 2]
Answer: (c) [-1/2, 1/2]
Explanation: Let, .
2x = sin θ.
Since, – 1 ≤ sin θ ≤ 1
– 1 ≤ 2x ≤ 1\Rightarrow
\sin^{-1}(2x)
\sin^{–1}x + \sin^{–1}y = \pi/2
\cos^{–1}x + \cos^{–1}y
\sin^{–1} x + \sin^{–1} y = \pi/2\Rightarrow [(\pi/2) – \cos^{-1}x] + [(\pi/2) – \cos^{-1}y] = \pi/2
\cos^{-1}x + \cos^{-1}y
\cos^{-1}x + \cos^{-1}y
\cos^{–1}x
\cos^{-1}x
\sin [\cot^{–1} (\cos (\tan^{–1} 1))]
\sin [\cot^{–1} (\cos (\tan^{–1} 1))]
\sin [\cot^{–1} (\cos (\tan^{–1} (\tan \pi/4))]
= \sin[\cot^{-1} (\cos \pi/4)]= \sin[\cot^{-1}(1/√2)]= \sin [\sin^{-1}(√(⅔))]
y = \cos^{–1} (x^2 – 4)
y = \cos^{–1} (x^2 – 4 )\Rightarrow \cos y = x^2 – 4
x^2 – 4
x^2
2 \sin^{–1}x + \cos^{–1}x
2 \sin^{–1}x + \cos^{–1}x
\sin^{–1}
\sin^{–1}x
\sin^{–1}x + (\sin^{–1}x + \cos^{–1}x)
2 \sin^{–1}x + \cos{–1}x
2 \sin^{–1}x + \cos{–1}x
\sin (2 \tan^{–1} (.75))
\sin 1.5
\sin (2\tan^{–1} (.75))
\tan^{–1} (.75)
\Rightarrow \tan \theta = 0.75\Rightarrow \tan \theta = 3/4
\sin (2\tan^{–1} (.75)) = \sin 2θ\theta
tan^{-1}(.75) = \theta
\sin (2\tan^{–1} (.75)) = .96
\sin(\tan^{-1} x)
x/√(1 – x^2)
1/√(1 – x^2)
1/√(1 + x^2)
x/√(1 + x^2)
x/√(1 + x^2)
\tan^{-1}x =
\sin \theta = x/√(1 + x^2)\cos \theta = 1/√(1 + x^2)
\sin(\tan^{-1} x) = \sin \theta = x/√(1 + x^2)
2\sec^{-1}2+\sin^{-1}(\frac{1}{2})
2\sec^{-1}2+\sin^{-1}(\frac{1}{2})= \frac{2\pi}{3} + \frac{\pi}{6}=\frac{5\pi}{6}
\tan^2(\sec^{-1}2) +\cot^2(\opratorname{cosec}^{-1}3)
f(x) =\sin^{-1}\sqrt{1-x}
\cot[\cos^{-1}(\frac{7}{25})]
\sin(\tan^{-1}x), |x|<1
\frac{x}{\sqrt{1-x^2}}
\frac{1}{\sqrt{1-x^2}}
\frac{1}{\sqrt{1+x^2}}
\frac{x}{\sqrt{1+x^2}}
\cos^{-1}\alpha+\cos^{-1}\beta+\cos^{-1}\gamma = 3\pi
\alpha(\beta +\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)
\cos^{-1}x
\sin^{-1}[\cos(33π/5)]
\cos^{-1}(2x-1)
\operatorname{cosec}^{-1}x
\cos(\sin^{-1}\frac{3}{5}+\cos^{-1})= 0
\cos^{-1}(2x^2-1),0 \leq x\lea 1
2\cos^{-1}x
\sin^{-1}x
\pi – 2\cos^{-1}x
\pi + 2\cos^{-1}x
\sqrt{1+\cos 2x} = \sqrt{2}\cos^{-1}(\cos x)
2\cos^{-1}(\frac{-1}{2})+2\sin^{-1}(\frac{-1}{2})-\cos^{-1}(-1)
\cot[\frac{1}{2}\sin^{-1}\frac{\sqrt{3}}{2}]
\sin(\cot^{-1}x)
\sqrt{1+x^2}
(1+x^2)^{-3/2}
(1+x^2)^{-1/2}$
Answer (d)
Class 12 matrix multiple choice question