Prove that $$cot^2\dfrac{\pi}{6}+cosec\dfrac{5\pi}{6}+3tan^2\dfrac{\pi}{6}=6$$

Asked by Pragya Singh | 1 year ago |  64

##### Solution :-

Substituting the values of $$cot\dfrac{\pi}{6},cosec \dfrac{5\pi}{6},tan\dfrac{\pi}{6}$$ on left hand side,

L.H.S. = $$(\sqrt{3})^2+cosec\dfrac{5\pi}{6}+3tan^2\dfrac{\pi}{6}$$

$$(\sqrt{3})^2+cosec(\pi-\dfrac{\pi}{6})+3(\dfrac{1}{\sqrt{3}})^2$$

$$3+cosec\dfrac{\pi}{6}+3\times \dfrac{1}{3}$$

Since cosecx repeat its value after interval of 2π,

We have, $$cosec\dfrac{5\pi}{6}=cosec\dfrac{\pi}{6}$$

L.H.S= 3+ 2 +1 =1

= R.H.S.

Hence proved.

Answered by Abhisek | 1 year ago

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