Question

Prove the following trigonometric identities:
cotθ-tanθ=2cos²θ-1/sinθcosθ

Answer 1

To Prove :

cotθ-tanθ=2cos²θ-1/sinθcosθ

• Using LHS ,

cotθ-tanθ = cosθ/sinθ- sinθ/cosθ

• Taking LCM ,

(cos²θ-sin²θ)/sinθcosθ

• putting sin²θ= 1- cos²θ

LHS = [cos²θ-(1-cos²θ)]/sinθcosθ

=2cos²θ-1/sinθcosθ

= RHS

hence proved

Answer 2

• cotθ-tanθ=2cos²θ-1/sinθcosθ
• L.H.S. = cotθ-tanθ
• using the formula, cotθ=cosθ/sinθ and tanθ=sinθ/cosθ we get,
• L.H.S.= cosθ/sinθ-sinθ/cosθ
• L.C.M. of the above is sinθcosθ,
• =(cos²θ-sin²θ)/sinθcosθ
• using the identity, sin²θ+cos²θ=1 and cos²θ=1-sin²θ, we get,
• L.H.S.= cos²θ-(1-cos²θ)/sinθcosθ
• L.H.S.=2cos²θ-1/sinθcosθ=R.H.S.