MCQ
$\tan \left[ {\frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{\sqrt 5 }}{3}} \right)} \right] = $
- ✓$\frac{{3 - \sqrt 5 }}{2}$
- B$\frac{{3 + \sqrt 5 }}{2}$
- C$\frac{2}{{3 - \sqrt 5 }}$
- Dએક પણ નહિ
Let $\frac{1}{2}{\cos ^{ - 1}}\frac{{\sqrt 5 }}{3} = \theta $
$\Rightarrow \cos 2\theta = \frac{{\sqrt 5 }}{3}$
But $\cos 2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} $
$\Rightarrow \frac{{\sqrt 5 }}{3} = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$
==>$\sqrt 5 + \sqrt 5 {\tan ^2}\theta = 3 - 3{\tan ^2}\theta $
==> $(\sqrt 5 + 3){\tan ^2}\theta = 3 - \sqrt 5 $
$\Rightarrow {\tan ^2}\theta = \frac{{3 - \sqrt 5 }}{{3 + \sqrt 5 }}$
==> ${\tan ^2}\theta = \frac{{{{(3 - \sqrt 5 )}^2}}}{4} $
$\Rightarrow \tan \theta = \frac{{3 - \sqrt 5 }}{2}$
On rationalising
==> $\tan \theta = \frac{{3 - \sqrt 5 }}{2} \times \frac{{3 + \sqrt 5 }}{{3 + \sqrt 5 }} = \frac{2}{{3 + \sqrt 5 }}$.
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વિધાન $1: $ $adj\left( {adj\;A} \right) = A$
વિધાન $2:$ $\left| {adj\;A} \right| = \left| A \right|$