Question

Prove that: cotθ- tanθ=(2cos^2θ-1)/(sinθcosθ).

Answer 1

Use the identities,

$2\cos^2 \theta-1=\cos 2\theta=\cos^2 \theta-\sin^2 \theta$.

Now,

$LHS=\cot \theta -\tan \theta =\frac{\cos \theta }{\sin \theta } -\frac{\sin \theta }{\cos \theta } \\ LHS=\cot \theta -\tan \theta =\frac{\cos^2 \theta-\sin^2 \theta }{\sin \theta \cos \theta } \\ LHS=\cot \theta -\tan \theta =\frac{2\cos^2 \theta-1 }{\sin \theta \cos \theta } =RHS$