How do you prove #(1-cot^2theta)/(1+cot^2theta) + 2cos^2theta = 1#?

2 Answers
Shwetank Mauria
Jun 7, 2016

Please see below.

Explanation:

#(1-cot^2theta)/(1+cot^2theta)+2cos^2theta#

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

= #((sin^2theta-cos^2theta)/sin^2theta)/((sin^2theta+cos^2theta)/sin^2theta)+2cos^2theta#

= #((sin^2theta-cos^2theta)/sin^2theta)xx(sin^2theta/(sin^2theta+cos^2theta))+2cos^2theta#

= #(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

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