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Inverse Trigonometric Functions

Multiple Choice Questions

271.

Given $0 \leq x \leq \frac{1}{2}$ , then the value of $\tan [\sin^{- 1} \{\frac{x}{\sqrt{2}} + \frac{\sqrt{1 - x^{2}}}{\sqrt{2}}\} - \sin^{- 1} (x)]$ is

1
$\sqrt{3}$
- 1
$\frac{1}{\sqrt{3}}$

272.

The value of $\sin (2 \sin^{- 1} (0.8))$ is equal to

0.48
$\sin (1.2 °)$
$\sin (1.6 °)$
0.96

273.

The value of $\sin^{- 1} (\frac{2 \sqrt{2}}{3}) + \sin^{- 1} (\frac{1}{3})$ is

$\frac{π}{4}$
$\frac{π}{6}$
$\frac{2 π}{3}$
$\frac{π}{2}$

274.

Solve for $x \tan^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} \tan^{- 1} (x) . x > 0$

- 1
$\sqrt{3}$
$\frac{1}{\sqrt{3}}$
1

275.

If $\frac{{(x + 1)}^{2}}{x^{3} + x} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1}$ , then $\csc^{- 1} (\frac{1}{A}) + \cot^{- 1} (\frac{1}{B}) + \sec^{- 1} (C)$ is equal to

$\frac{π}{6}$
$\frac{5 π}{6}$
$\frac{π}{2}$
0

276.

If $α \leq 2 \sin^{- 1} (x) + \cos^{- 1} (x) \leq β$ , then

$α = 0, β = π$
$α = \frac{- π}{2}, β = \frac{π}{2}$
$α = 0, β = 2 π$
$α = \frac{- π}{2}, β = \frac{3 π}{2}$

277.
The value of $\sin^{- 1} (\cos (\frac{53 π}{5}))$ is

$\frac{3 π}{5}$

$\frac{- 3 π}{5}$

$\frac{π}{10}$

$\frac{- π}{10}$

$\frac{- π}{10}$

$We have, \sin^{- 1} (\cos (\frac{53 π}{5})) = \sin^{- 1} [\cos (10 π + \frac{3 π}{5})] = \sin^{- 1} [\cos (\frac{3 π}{5})] = \sin^{- 1} [\sin (\frac{π}{2} - \frac{3 π}{5})] = \sin^{- 1} [\sin (\frac{5 π - 6 π}{10})] = \sin^{- 1} [\sin (\frac{- π}{10})] = \frac{- π}{10}$