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

Asked by Pragya Singh | 1 year ago |  96

##### Solution :-

Substituting the values of $$sin\dfrac{3\pi}{4},cos \dfrac{\pi}{4},sec\dfrac{\pi}{3}$$ on left hand side,

L.H.S.= $$2 sin^2\dfrac{3\pi}{4}+2cos^2\dfrac{\pi}{4}+2sec^2\dfrac{\pi}{3}$$

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

$$2\{sin\dfrac{\pi}{4}\}^2+2\times \dfrac{1}{2}+8$$

Since sinx repeat its value after interval of 2π

We have, $$sin\dfrac{3\pi}{4}= sin\dfrac{\pi}{4}$$

L.H.S 1 +1+ 8

=10 =  R.H.S.

Hence proved.

Answered by Abhisek | 1 year ago

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