MCQ
${\cot ^{ - 1}}\left[ {\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}} \right] = $
- A$\pi - x$
- B$2\pi - x$
- C$\frac{x}{2}$
- ✓$\pi - \frac{x}{2}$
$ = {\cot ^{ - 1}}\left[ {\frac{{(\sqrt {1 - \sin x} + \sqrt {1 + \sin x} )}}{{(\sqrt {1 - \sin x} - \sqrt {1 + \sin x} )}}.\frac{{(\sqrt {1 - \sin x} + \sqrt {1 + \sin x} )}}{{(\sqrt {1 - \sin x} + \sqrt {1 + \sin x} )}}} \right]$
$= {\cot ^{ - 1}}\left[ {\frac{{(1 - \sin x) + (1 + \sin x) + 2\sqrt {1 - {{\sin }^2}x} }}{{(1 - \sin x) - (1 + \sin x)}}} \right]$
$= {\cot ^{ - 1}}\left[ {\frac{{2(1 + \cos x)}}{{ - 2\sin x}}} \right] = {\cot ^{ - 1}}\left[ { - \frac{{2{{\cos }^2}(x/2)}}{{2\sin (x/2)\cos (x/2)}}} \right]$
$= {\cot ^{ - 1}}\left( { - \cot \frac{x}{2}} \right) = {\cot ^{ - 1}}\left[ {\cot \left( {\pi - \frac{x}{2}} \right)} \right] = \pi - \frac{x}{2}$.
Trick : Put $x = \frac{\pi }{4}$, so that the expression becomes
${\cot ^{ - 1}}\left[ {\frac{{\sqrt {\sqrt 2 - 1} + \sqrt {\sqrt 2 + 1} }}{{\sqrt {\sqrt 2 - 1} - \sqrt {\sqrt 2 + 1} }}} \right]$
$= {\cot ^{ - 1}}\left[ {\frac{{\sqrt 2 - 1 + \sqrt 2 + 1 + 2\sqrt {2 - 1} }}{{\sqrt 2 - 1 - \sqrt 2 - 1}}} \right]$
$= {\cot ^{ - 1}}\left[ {\frac{{2\sqrt 2 + 2}}{{ - 2}}} \right] = {\cot ^{ - 1}}( - 1 - \sqrt 2 ) = 157.5^\circ $.
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વિધાન ${\text{ - 2 : }}$ રેખા $\frac{{\text{x}}}{{\text{1}}}\,\, = \,\,\frac{{y\,\, - \,\,1}}{2}\,\, = \,\,\frac{{z\,\, - \,\,2}}{3}\,$ એ $A\,\,\left( {1,\,\,0,\,\,7} \right)$ અને $B\,\,\left( {1,\,\,6,\,3} \right)$ ને જોડતા રેખાખડને લંબ-દ્રીભાજે છે