Prove that $$sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{3}-tan^2\dfrac{\pi}{4}=-\dfrac{1}{2}$$

Asked by Pragya Singh | 2 years ago |  78

##### Solution :-

Substituting the values of $$sin\dfrac{\pi}{6},cos \dfrac{\pi}{6},tan\dfrac{\pi}{4}$$ on left hand side,

$$sin^2\dfrac{\pi}{6}+cos^2 \dfrac{\pi}{6}-tan^2\dfrac{\pi}{4}$$

$$( \dfrac{1}{2})^2+(\dfrac{1}{2})^2-(1)^2$$

$$\dfrac{1}{4}+\dfrac{1}{4}-1$$

$$- \dfrac{1}{2}= R.H.S.$$

Hence proved.

Answered by Abhisek | 2 years ago

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