Question

311 cos cos 211 + cot cot (211 + X) =

Answer 1

The given trigonometric function is cos( 3π 2 +x )cos( 2π+x )[ cot( 3π 2 −x )+cot( 2π+x ) ]=1.

Solve the left hand side of the trigonometric function.

L.H.S.=cos( 3π 2 +x )cos( 2π+x )[ cot( 3π 2 −x )+cot( 2π+x ) ] =cos( π+( π 2 +x ) )cosx[ cot( π+( π 2 −x ) )+cotx ]       ∵cos( π+θ )=−cosθ;cot( π+θ )=cotθ =−cos( π 2 +x )cosx[ cot( π 2 −x )+cotx ]                       ∵cos( π 2 +θ )=−sinθ;cot( π 2 −θ )=tanθ =sinxcosx[ tanx+cotx ]

Further simplify the above expression.

L.H.S.=sinxcosx[ tanx+cotx ] =sinxcosx[ sinx cosx + cosx sinx ] = sin 2 x+ cos 2 x =1

This is equal to the right hand side of the given trigonometric expression.

Thus, the trigonometric function is cos( 3π 2 +x )cos( 2π+x )[ cot( 3π 2 −x )+cot( 2π+x ) ]=1.

Since L.H.S. = R.H.S.

Hence proved.