\cot \theta-\tan \theta=\frac{2 \cos ^{2} \theta-1}{\operatorname{sen}

1.	\cot \theta-\tan \theta=\frac{2 \cos ^{2} \theta-1}{\operatorname{sen} \theta \cos \theta}
Answer» (2cos²θ - 1)/(sinθcosθ) = cotθ - tanθ LHS =(2cos²θ - cos²θ - sin²θ)/(sinθcosθ) ∵ 1 = cos²θ + sin²θ (cos²θ - sin²θ)/ sinθcosθ Dividing throughout by cos²θ = (1 - tan²θ)/tanθ = cot θ - tanθ = RHS Hence Proved

\cot \theta-\tan \theta=\frac{2 \cos ^{2} \theta-1}{\operatorname{sen} \theta \cos \theta}

Answer»

(2cos²θ - 1)/(sinθcosθ) = cotθ - tanθ

LHS

=(2cos²θ - cos²θ - sin²θ)/(sinθcosθ)

∵ 1 = cos²θ + sin²θ

(cos²θ - sin²θ)/ sinθcosθ

Dividing throughout by cos²θ

= (1 - tan²θ)/tanθ

= cot θ - tanθ

= RHS

Hence Proved