MCQ
$\sin [{\cot ^{ - 1}}(\cos {\tan ^{ - 1}}x)] =$
- A$\frac{x}{{\sqrt {{x^2} + 2} }}$
- B$\frac{x}{{\sqrt {{x^2} + 1} }}$
- C$\frac{1}{{\sqrt {{x^2} + 2} }}$
- ✓$\sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}} $
✓
Answer
Correct option: D.
$\sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}} $
d
(d) $\sin \,[{\cot ^{ - 1}}\,(\cos \,\,{\tan ^{ - 1}}x)]$
(d) $\sin \,[{\cot ^{ - 1}}\,(\cos \,\,{\tan ^{ - 1}}x)]$
$ = \sin \,\left[ {{{\cot }^{ - 1}}\,\left( {\cos \,\,{{\cos }^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}} \right)} \right]$
$ = \sin \,\left[ {{{\cot }^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}} \right] = \sin \,\left[ {{{\sin }^{ - 1}}\sqrt {\frac{{1 + {x^2}}}{{2 + {x^2}}}} } \right]$
$ = \sqrt {\frac{{1 + {x^2}}}{{2 + {x^2}}}} $.
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