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प्रश्न
2 tan−1 {cosec (tan−1 x) − tan (cot−1 x)} is equal to
पर्याय
cot−1 x
cot−1`1/x`
tan−1 x
none of these
उत्तर
(c) tan−1 x
Let `tan^-1x=y`
So, `x=tany`
\[\therefore 2 \tan^{- 1} \left\{ cosec\left( \tan^{- 1} x \right) - \tan\left( co t^{- 1} x \right) \right\} = 2 \tan^{- 1} \left\{ cosec\left( \tan^{- 1} x \right) - \tan\left( \tan^{- 1} \frac{1}{x} \right) \right\} \]
\[ = 2 \tan^{- 1} \left\{ cosec\left( \tan^{- 1} x \right) - \frac{1}{x} \right\}\]
\[ = 2 \tan^{- 1} \left\{ cosec {y} - \frac{1}{\tan{y}} \right\}\]
\[ = 2 \tan^{- 1} \left\{ \frac{1 - \cos{y}}{\sin{y}} \right\}\]
\[ = 2 \tan^{- 1} \left\{ \frac{2 \sin^2 \frac{y}{2}}{\sin{y}} \right\} \]
\[ = 2 \tan^{- 1} \left\{ \frac{2 \sin^2 \frac{y}{2}}{2\sin\frac{y}{2}\cos\frac{y}{2}} \right\}\]
\[ = 2 \tan^{- 1} \left\{ \tan\frac{y}{2} \right\}\]
\[ = y\]
\[ = \tan^{- 1} x\]
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