Question

सिद्ध कीजिए: $2\sin^2 \dfrac {3\pi}{4} + 2\cos^2 \dfrac{\pi}{4} + 2\sec^2 \dfrac{\pi}{3} = 10$

Answer 1

सिद्ध कीजिए: $2\sin^2 \dfrac {3\pi}{4} + 2\cos^2 \dfrac{\pi}{4} + 2\sec^2 \dfrac{\pi}{3} = 10$

हम जानते हैं कि

${sin}^{2} \theta + {cos}^{2} \theta = 1 \\ \\ sin(\pi - \theta) = sin \: \theta \\ \\{sec \frac({\pi}{3} })=2$
$sin^2 (\dfrac {3\pi}{4}) = > {sin}^{2} (\pi - \frac{\pi}{4}) \\ \\ = > {sin}^{2} ( \frac{\pi}{4} )$
$2\sin^2 \dfrac {\pi}{4} + 2\cos^2 \dfrac{\pi}{4} + 2\sec^2 \dfrac{\pi}{3} \\ \\ 2( {sin}^{2} \frac{\pi}{4} + {cos}^{2} \frac{\pi}{4} ) + 2 {sec}^{2} \frac{\pi}{3} \\ \\ = > 2 + 2 {(sec \frac{\pi}{3} })^{2} \\ \\ = > 2 + 2( {2)}^{2} \\ \\ = > 2 + 2 \times 4 \\ \\ = > 2 + 8 \\ \\ = > 10 \\ \\ = > RHS$

Answer 2

Answer:

Step-by-step explanation:

$2\sin^2 \dfrac {3\pi}{4} + 2\cos^2 \dfrac{\pi}{4} + 2\sec^2 \dfrac{\pi}{3} = 10$

L.H.S.  

     =   $2\sin^2 \dfrac {3\pi}{4} + 2\cos^2 \dfrac{\pi}{4} + 2\sec^2 \dfrac{\pi}{3}$

    =   $2 sin^{2} ( \pi -\frac{\pi }{4} )+ 2(\frac{1}{\sqrt{2} } )^{2}+ 2(2)^{2}$

  =   $2 sin^{2} ( \frac{\pi }{4} )+ 2*\frac{1}{2} + 2*4$

  = $2 sin^{2} ( \frac{\pi }{4} )+ 1 + 8$

   =  $2( \frac{1}{\sqrt{2} } )^{2}+ 1 + 8$

   =  $2 * \frac{1}{2} +1+8$

   =  10  =  R.H.S.

अतः   L.H.S.  =  R.H.S.