MCQ
જો $\tan (x + y) = 33$ અને $x = {\tan ^{ - 1}}3,$ તો $y = . . .$
- A$0.3$
- B${\tan ^{ - 1}}(1.3)$
- ✓${\tan ^{ - 1}}(0.3)$
- D${\tan ^{ - 1}}\left( {\frac{1}{{18}}} \right)$
$\Rightarrow y = {\tan ^{ - 1}}33 - {\tan ^{ - 1}}3$
$ = {\tan ^{ - 1}}\frac{{33 - 3}}{{1 + 99}} = {\tan ^{ - 1}}\frac{{30}}{{100}}$
==> $y = {\tan ^{ - 1}}(0.3)$.
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