$=$$\sin {\frac{{165}}{2}^0}\,.\,\cos {\frac{{75}}{2}^0}$
$=$ $\frac{1}{2}\left[ {\sin {{120}^0}\, + \,\sin {{45}^0}} \right]$
$=$ $\frac{{\sqrt 6 \, + 1}}{{4\sqrt 2 }}$
$B$ $=$$\sin 127{\frac{1}{2}^0}\,.\sin 97{\frac{1}{2}^0}$ $=$$\frac{1}{2}\,\left[ {\cos {{30}^0}\, - \,\cos {{225}^0}} \right]$ $=$$\frac{1}{2}\,\left[ {\frac{{\sqrt 3 }}{2}\, + \,\frac{1}{{\sqrt 2 }}} \right]$ $=$$\frac{{\sqrt 6 \, + 2}}{{4\sqrt 2 }}$ $= \frac{{\sqrt 3 + \sqrt 2 }}{4}$
$\Rightarrow\, A = B \,\Rightarrow True $
$[B]$ $tan(A-B) =$$\frac{{\tan A\, - \,\tan B}}{{1 + \tan A\,\,\tan B}}$
$=$$\frac{{\frac{{\sqrt 3 }}{{4 - \sqrt 3 }}\, - \,\frac{{\sqrt 3 }}{{4 + \sqrt 3 }}}}{{1 + \frac{{\sqrt 3 \,.\,\sqrt 3 }}{{\left( {4 - \sqrt 3 } \right)\,\left( {4 + \sqrt 3 } \right)}}}}$
$=$$\frac{{\sqrt 3 \,\left[ {4 + \sqrt 3 - 4 + \sqrt 3 } \right]}}{{16 - 3 + 3}}$
$= 3/8$ $\Rightarrow$ rational
$[C]$ $sin2 = + ; sin3 = + ; sin5 = -$
$[D]$ $ sin2\theta = \frac{{1 \pm \,\sqrt 5 }}{2}$
$\Rightarrow sin2\theta =$ $\frac{{1 - \sqrt 5 }}{2}$ (not possible)
$sin2\theta = \frac{{1 + \,\sqrt 5 }}{2}\,\,\, > \,\,1$ not possible
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કારણ $(R)$ : આપેલ બિંદુથી રેખા સુધીની બીજગણિતીય લંબ અંતર વિરૂદ્ધ ચિહ્ન ધરાવે છે.