MCQ
$2{\tan ^{ - 1}}(\cos x) = {\tan ^{ - 1}}({\rm{cose}}{{\rm{c}}^2}x),$ તો $ x =$
- A$\frac{\pi }{2}$
- B$\pi $
- C$\frac{\pi }{6}$
- ✓$\frac{\pi }{3}$
==> ${\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{1 - {{\cos }^2}x}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{{{{\sin }^2}x}}} \right)$
$ \Rightarrow \frac{{2\cos x}}{{{{\sin }^2}x}} = \frac{1}{{{{\sin }^2}x}}$
==> $2\cos x = 1$
$ \Rightarrow x = \frac{\pi }{3}$.
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$ \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R} $
$ \mathrm{L}_2: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{k}, \mu \in \mathbb{R}$
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