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Inverse Trigonometric Functions

Multiple Choice Questions

321.

${sech}^{- 1} (\frac{1}{2}) - {cosec}^{- 1} (\frac{3}{4}) = ?$

$\log_{e} (3 (2 + \sqrt{3}))$
$\log_{e} (\frac{1 + \sqrt{3}}{3})$
$\log_{e} (\frac{2 + \sqrt{3}}{3})$
$\log_{e} (\frac{2 - \sqrt{3}}{3})$

322.

If $\sin^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . . \infty) + \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . \infty) = \frac{π}{2}$ and 0 < x < $\sqrt{2}$ , then x is equal to

$\frac{1}{2}$
1
$- \frac{1}{2}$
- 1

323.
If $2 \sinh^{- 1} (\frac{a}{\sqrt{1 - a^{2}}}) = \log (\frac{1 + x}{1 - x})$ , then x is equal to

a

$\frac{1}{a}$

$\sqrt{1 - a^{2}}$

$\frac{1}{\sqrt{1 - a^{2}}}$

$Given that, 2 \sinh^{- 1} (\frac{a}{\sqrt{1 - a^{2}}}) = \log (\frac{1 + x}{1 - x}) \Rightarrow 2 \sinh^{- 1} (x) = 2 \log (x + \sqrt{1 + x^{2}}) [∵ \sinh^{- 1} (x) = \log (x + \sqrt{1 + x^{2}})] = \log {(x + \sqrt{1 + x^{2}})}^{2} = \log (x^{2} + 1 + x^{2} + 2 x \sqrt{1 + x^{2}}) 2 \sinh^{- 1} (x) = \log (2 x^{2} + 1 + 2 x \sqrt{1 + x^{2}}) = \log [\frac{2 a}{1 - a^{2}} + 1 + 2 \times \frac{a}{\sqrt{1 - a^{2}}} \times \sqrt{1 + \frac{a^{2}}{1 - a^{2}}}] [∵ x = \frac{a}{\sqrt{1 - a^{2}}}] = \log [\frac{2 a}{1 - a^{2}} + 1 + \frac{2 a}{\sqrt{1 - a^{2}}} \times \frac{1}{\sqrt{1 - a^{2}}}] = \log [\frac{2 a}{1 - a^{2}} + 1 + \frac{2 a}{1 - a^{2}}] = \log [\frac{2 a^{2} + 1 - a^{2} + 2 a}{1 - a^{2}}] = \log [\frac{a^{2} + 2 a + 1}{1 - a^{2}}]$

$Put the value of 2 \sinh^{- 1} (\frac{a}{\sqrt{1 - a^{2}}}), we get \log [\frac{a^{2} + 2 a + 1}{1 - a^{2}}] = \log [\frac{1 + x}{1 - x}] \Rightarrow \frac{a^{2} + 2 a + 1}{1 - a^{2}} = \frac{1 + x}{1 - x} \Rightarrow \frac{a^{2} + 2 a + 1 + 1 - a^{2}}{a^{2} + 2 a + 1 - 1 + a^{2}} = \frac{1 + x + 1 - x}{1 - x - 1 + x} \Rightarrow \frac{2 + 2 a}{2 a^{2} + 2 a} = \frac{1}{x} [∵ by componendo and divdendo rule] \Rightarrow \frac{2 (1 + a)}{2 a (1 + a)} = \frac{1}{x} \Rightarrow \frac{1}{a} = \frac{1}{x} \Rightarrow x = a$