Question

सिद्ध कीजिए: $2\sin^2 \dfrac {\pi}{6} + cosec^2 \dfrac{7\pi}{6} \cos^2 \dfrac{\pi}{3} = \dfrac{3}{2}$

Answer 1

Answer:

Step-by-step explanation:

$2sin^{2}\frac{\pi }{6} + cosec^{2}\frac{7\pi }{6} cos^{2}\frac{\pi }{3}$=3/2

L.H.S.

   = $2sin^{2}\frac{\pi }{6} + cosec^{2}\frac{7\pi }{6} cos^{2}\frac{\pi }{3}$

   = $2sin^{2}\frac{\pi }{6} + cosec^{2}(\pi +\frac{\pi}{6}) cos^{2}\frac{\pi }{3}\\\\=2sin^{2}\frac{\pi }{6} + (-cosec\frac{\pi }{6} )^{2} . cos^{2}\frac{\pi }{3} \\\\= 2sin^{2}\frac{\pi }{6} + cosec^{2}\frac{\pi }{6} . cos^{2}\frac{\pi }{3}$

  =  $2 sin^{2} 30' + cosec^{2} 30' . cos^{2} 60'$

  =  2. $(\frac{1}{2}) ^{2} + (2^{2}) . (\frac{1}{2}) ^{2}$

  =  2×$\frac{1}{4}+4*\frac{1}{4}$

  =  $\frac{1}{2} +1= \frac{3}{2}$

  =  R.H.S.