MCQ
$cos\left\{tan^{-1}\left(tan \frac{15\pi}{4}\right)\right\}=........$
- A$1$
- B$-\frac{1}{\sqrt{2}}$
- ✓$\frac{1}{\sqrt{2}}$
- Dએક પણ નહી.
$cos(tan^{-1}(tan\frac{15\pi}{4})$
અહી $tan^{-1}(tanx)=x$ થાય
$tan\frac{{15\pi }}{4} = tan\left( {4\pi - \frac{\pi }{4}} \right) = \left( {tan\frac{{ - \pi }}{4}} \right)$
$\therefore cos(tan^{-1}(tan(\frac{-\pi}{4})$
$=cos(\frac{-\pi}{4})$
$=\frac{1}{\sqrt{2}}$
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| $x=x$ | $0$ | $1$ | $2$ | $3$ |
| $p(x)$ | $\frac{1}{4}$ | $\frac{1}{8}$ | $\frac{1}{2}$ | $\frac{1}{8}$ |