Prove that, $$cos(\dfrac{3\pi}{2}+x)(Cos 2π+x)[cot(\dfrac{3\pi}{2}-x)+(Cot 2π+x)]=1$$

Asked by Abhisek | 1 year ago |  115

##### Solution :-

We know that cotx repeats same value after an interval 2π

L.H.S.= $$cos(\dfrac{3\pi}{2}+x)(Cos 2π+x)$$

$$[cot(\dfrac{3\pi}{2}-x)+(Cot 2π+x)]=1$$

=sinxcosx[tanx+cotx]

Substituting $$tanx=\dfrac{sinx}{cosx}$$ and

$$cotx=\dfrac{cosx}{sinx}$$,

L.H.S=sinxcosx $$(\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx})$$

$$(sinxcosx) [\dfrac{sin^2x+cos^2x}{sinxcosx}]$$

= 1

= R.H.S.

Hence proved.

Answered by Pragya Singh | 1 year ago

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