Trig Identities (1+tan^2θ)/(csc^2θ) = tan^2θ?
(1+tan^2θ)/(csc^2θ) = tan^2θ
I havn't done any like this. Can someone help me out?
More Questions From This User See All
(1+tan^2θ)/(csc^2θ) = tan^2θ
I havn't done any like this. Can someone help me out?
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
1+tan^2θ=sec^2θ
take LHS
(1+tan^2θ)/csc^2θ
=sec^2θ/csc^2θ
=sin^2θ/cos^2θ
=tan^2θ
LHS=RHS
1+tan^2O =sec^2O
sec^2O=1/cos^2O
csc^2O=1/sin^2O
sec^2O/csc^2O=sin^2O/cos^2O=tan^2O