MCQ
જો ${\cot ^{ - 1}}x + {\tan ^{ - 1}}3 = \frac{\pi }{2}$, તો $x =$
- A$1/3$
- B$1/4$
- ✓$3$
- D$4$
$ \Rightarrow \,\,{\cot ^{ - 1}}x + {\tan ^{ - 1}}3 = \frac{\pi }{2}\,\, $
$\Rightarrow \,\,{\tan ^{ - 1}}\frac{1}{x} + {\tan ^{ - 1}}3 = \frac{\pi }{2}$
$ \Rightarrow \,\,{\tan ^{ - 1}}\left( {\frac{{\frac{1}{x} + 3}}{{1 - \frac{1}{x}.3}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{0}} \right)$
$ \Rightarrow \,\,\frac{{3x + 1}}{{x - 3}} = \frac{1}{0}\,\, \Rightarrow \,\,x = 3$
Aliter : As we know that, ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2},$
therefore for the given question, $ x$ should be $3.$
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વિધાન $-2$ : સમીકરણ કે જે $\alpha $ સ્વરૂપ માં છે
$\left| {\begin{array}{*{20}{c}}
{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\
{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\
{\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha }
\end{array}} \right| = 0$
નું એક માત્ર બીજ અંતરાલ $\left( {0\,,\,\frac{\pi }{2}} \right)$ માં છે .