How do you prove (1-cot^2theta)/(1+cot^2theta) + 2cos^2theta = 11−cot2θ1+cot2θ+2cos2θ=1?

2 Answers
Shwetank Mauria
Jun 7, 2016

Please see below.

Explanation:

(1-cot^2theta)/(1+cot^2theta)+2cos^2theta1−cot2θ1+cot2θ+2cos2θ

= (1-cos^2theta/sin^2theta)/(1+cos^2theta/sin^2theta)+2cos^2theta1−cos2θsin2θ1+cos2θsin2θ+2cos2θ

= ((sin^2theta-cos^2theta)/sin^2theta)/((sin^2theta+cos^2theta)/sin^2theta)+2cos^2thetasin2θ−cos2θsin2θsin2θ+cos2θsin2θ+2cos2θ

= ((sin^2theta-cos^2theta)/sin^2theta)xx(sin^2theta/(sin^2theta+cos^2theta))+2cos^2theta(sin2θ−cos2θsin2θ)×(sin2θsin2θ+cos2θ)+2cos2θ

= (sin^2theta-cos^2theta)/cancel(sin^2theta)xxcancel(sin^2theta)/1+2cos^2theta

= sin^2theta-cos^2theta+2cos^2theta

= sin^2theta+cos^2theta=1

Joel Kindiak
Jun 7, 2016

Use trigonometric identities. Proof-writing is a skill, not content-based question.

Explanation:

LHS
=(1-cot^2 theta)/(1+cot^2 theta) + 2 cos^2 theta
=(1-cot^2 theta)/(csc^2 theta) + 2 cos^2 theta
=1/(csc^2 theta) - (cot^2 theta)/(csc^2 theta) + 2 cos^2 theta
=sin^2 theta - (cos^2 theta)/(sin^2 theta) * sin^2 theta + 2 cos^2 theta
=sin^2 theta - cos^2 theta + 2 cos^2 theta
=sin^2 theta + cos^2 theta
=1
= RHS

Related questions
Impact of this question
7204 views around the world
You can reuse this answer
Creative Commons License