MCQ
$2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = $
- A${\tan ^{ - 1}}\left( {\frac{{49}}{{29}}} \right)$
- B$\frac{\pi }{2}$
- C$0$
- ✓$\frac{\pi }{4}$
$ = {\tan ^{ - 1}}\frac{3}{4} + {\tan ^{ - 1}}\frac{1}{7} = {\tan ^{ - 1}}\left( {\frac{{(3/4) + (1/7)}}{{1 - (3/4) \times (1/7)}}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{25}}{{25}}} \right) = \frac{\pi }{4}$.
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$P = \left\{ {\left( {a,b} \right):{{\sec }^2}\,a - {{\tan }^2}\,b = 1\,} \right\}$. Then $P$ is
$\Big(\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}}\Big)$
$\big(2\sqrt{3},-2\sqrt{3},2\sqrt{3}\big)$
$-\big(2\sqrt{3},-2\sqrt{3},2\sqrt{3}\big)$
$-\big(6\sqrt{3},-6\sqrt{3},6\sqrt{3}\big)$