MCQ
$\cos \left[ {{{\tan }^{ - 1}}\frac{1}{3} + {{\tan }^{ - 1}}\frac{1}{2}} \right] = $
- ✓$\frac{1}{{\sqrt 2 }}$
- B$\frac{{\sqrt 3 }}{2}$
- C$\frac{1}{2}$
- D$\frac{\pi }{4}$
$ = \cos \,\{ {\tan ^{ - 1}}(1)\} = \cos \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}$.
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$\left[\begin{array}{ccc}1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}1 \\ -1 \\ 1\end{array}\right]$
of linear equations, has infinitely many solutions, then $1+\alpha+\alpha^2=$
$STATEMENT -1$ : For each real $\mathrm{t}$, there exists a point $\mathrm{c}$ in $[\mathrm{t}, \mathrm{t}+\pi]$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$. because
$STATEMENT -2$: $f(t)=f(t+2 \pi)$ for each real $t$.