MCQ
$\sin \left( {\frac{1}{2}{{\cos }^{ - 1}}\frac{4}{5}} \right) = $
- ✓$\frac{1}{{\sqrt {10} }}$
- B$ - \frac{1}{{\sqrt {10} }}$
- C$\frac{1}{{10}}$
- D$ - \frac{1}{{10}}$
Now $\sin \left( {\frac{1}{2}{{\cos }^{ - 1}}\frac{4}{5}} \right) = \sin \left( {\frac{x}{2}} \right)$ .....$(ii)$
From $(i),$ $\cos x = \frac{4}{5}$
==> $1 - 2{\sin ^2}\frac{x}{2} = \frac{4}{5}$
==> $2{\sin ^2}\frac{x}{2} = 1 - \frac{4}{5} = \frac{1}{5}$
$ \Rightarrow \sin \frac{x}{2} = \sqrt {\frac{1}{{10}}} $.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
If
$\text{P}(\text{A})=\frac{2}{5},\text{P}(\text{B})=\frac{3}{5}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{5},$ then $\text{P}\Big(\frac{\text{A}'}{\text{B}'}\Big)\cdot\text{P}\Big(\frac{\text{B}'}{\text{A}'}\Big)$ is equas: