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Inverse Trigonometric Functions

Multiple Choice Questions

111.

If $\cos^{- 1} (p) + \cos^{- 1} (q) + \cos^{- 1} (r) = π$ , then p² + q² + r² + 2pqr is equal to

112.

If $\sin^{- 1} (\frac{x}{5}) + \csc^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then x is equal to

113.

$\sin (\frac{1}{2} \cos^{- 1} (\frac{4}{5}))$

$- \frac{1}{\sqrt{10}}$
$\frac{1}{\sqrt{10}}$
$- \frac{1}{10}$
$\frac{1}{10}$

114.

If $\cos^{- 1} (x) + \cos^{- 1} (y) + \cos^{- 1} (z) = 3 π$ , then xy + yz + zx is equal to

115.

If $\tan^{- 1} (\frac{x + 1}{x - 1}) + \tan^{- 1} (\frac{x - 1}{x}) = \tan^{- 1} (- 7)$ , then the value of x is

zero
- 2
1
2

116.

If $\cos^{- 1} (\sqrt{p}) + \cos^{- 1} (\sqrt{1 - p}) + \cos^{- 1} (\sqrt{1 - q}) = \frac{3 π}{4}$ , then the value of q is

$\frac{2}{\sqrt{2}}$
1
1/2
1/3

117.
The soluton of $\sin^{- 1} (x) - \sin^{- 1} (2 x) = \pm \frac{π}{3}$ is

$\pm \frac{1}{3}$

$\pm \frac{1}{4}$

$\pm \frac{\sqrt{3}}{2}$

$\pm \frac{1}{2}$

$\pm \frac{1}{2}$

$\sin^{- 1} (x) - \sin^{- 1} (2 x) = \pm \frac{π}{3} \Rightarrow \sin^{- 1} (x) - \sin^{- 1} (\pm \frac{\sqrt{3}}{2}) = \sin^{- 1} (2 x) \Rightarrow \sin^{- 1} [x \sqrt{1 - \frac{3}{4}} - (\pm \frac{\sqrt{3}}{2} \sqrt{1 - x^{2}})] = \sin^{- 1} (2 x) \Rightarrow \frac{x}{2} - (\pm \frac{\sqrt{3}}{2} \sqrt{1 - x^{2}}) = 2 x \Rightarrow - (\pm \sqrt{3} \sqrt{1 - x^{2}}) = 3 x On squaring both sides, we get 3 (1 - x^{2}) = 9 x^{2} \Rightarrow 1 - x^{2} = 3 x^{2} \Rightarrow 4 x^{2} = 1 \Rightarrow x = \pm \frac{1}{2}$

118.

If $\cos^{- 1} (x) = α, (0 < x < 1)$ and $\sin^{- 1} (2 x \sqrt{1 - x^{2}})$ + $\sec^{- 1} (\frac{1}{2 x^{2} - 1}) = \frac{2 π}{3}$ , then $\tan^{- 1} (2 x)$ equals :