MCQ
If $\tan ({\cos ^{ - 1}}x) = \sin \left( {{{\cot }^{ - 1}}\frac{1}{2}} \right)$, then $x =$
- A$ \pm \frac{5}{3}$
- ✓$ \pm \frac{{\sqrt 5 }}{3}$
- C$ \pm \frac{5}{{\sqrt 3 }}$
- DNone of these
Let ${\cot ^{ - 1}}\frac{1}{2} = \phi \Rightarrow \frac{1}{2} = \cot \phi $
$ \Rightarrow \sin \phi = \frac{1}{{\sqrt {1 + {{\cot }^2}\phi } }} = \frac{2}{{\sqrt 5 }}$
Let ${\cos ^{ - 1}}x = \theta \Rightarrow \sec \theta = \frac{1}{x} $
$\Rightarrow \tan \theta = \sqrt {{{\sec }^2}\theta - 1} $
$ \Rightarrow \tan \theta = \sqrt {\frac{1}{{{x^2}}} - 1} $
$\Rightarrow \tan \theta = \frac{{\sqrt {1 - {x^2}} }}{x}$
So, $\tan \{ {\cos ^{ - 1}}(x)\} = \sin \left( {{{\cot }^{ - 1}}\frac{1}{2}} \right)$
$ \Rightarrow \tan \left( {{{\tan }^{ - 1}}\frac{{\sqrt {1 - {x^2}} }}{x}} \right) = \sin \left( {{{\sin }^{ - 1}}\frac{2}{{\sqrt 5 }}} \right)$
$ \Rightarrow \frac{{\sqrt {1 - {x^2}} }}{x} = \frac{2}{{\sqrt 5 }} \Rightarrow \sqrt {(1 - {x^2})5} = 2x$
Squaring both sides, we get $x = \pm \frac{{\sqrt 5 }}{3}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
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