MCQ
${\tan ^{ - 1}}\frac{{1 - {x^2}}}{{2x}} + {\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} = $
- A$\frac{\pi }{4}$
- ✓$\frac{\pi }{2}$
- C$\pi $
${\tan ^{ - 1}}\frac{{1 - {x^2}}}{{2x}} + {\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}}$
$ = {\tan ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{2\tan \theta }}} \right) + {\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$
$ = {\tan ^{ - 1}}(\cot 2\theta ) + {\cos ^{ - 1}}(\cos 2\theta )$
$ = \frac{\pi }{2} - 2\theta + 2\theta = \frac{\pi }{2}$.
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$(A)$ $1$ $(B)$ $2$ $(C)$ $3$ $(D)$ $4$
| $I$ | $II$ | $III$ | $IV$ |
| $f'(x) = \frac{9}{{28}} x^{7/3} +9$ | $f (x) = \frac{9}{{28}} x^{7/3} -2$ | $f (x) = \frac{3}{{4}}\,x^{4/3} +6$ | $f'(x) =\frac{3}{{4}}\,x^{4/3} -4$ |