Question

Prove the identity √1 - tan²∅ ÷ cot²∅ - 1 = tan∅​

Answer 1

Answer:

The given equation is:

\frac{1+tan^2{\theta}}{1+cot^2{theta}}=(\frac{1-tan{\theta}}{1-cot{\theta}})^{2}1+cot2theta1+tan2θ=(1−cotθ1−tanθ)2

Taking the LHS of the above equation, we have

\frac{1+tan^2{\theta}}{1+cot^2{theta}}1+cot2theta1+tan2θ

=\frac{sec^2\theta}{cosec^2\theta}cosec2θsec2θ

=\frac{sin^2\theta}{cos^2\theta}cos2θsin2θ

=tan^2{\theta}tan2θ

Now, taking the RHS of the above equation, we have

(\frac{1-tan{\theta}}{1-cot{\theta}})^{2}(1−cotθ1−tanθ)2

=(\frac{1-tan{\theta}}{1-\frac{1}{tan\theta}})^{2}(1−tanθ11−tanθ)2

=(\frac{(1-tan{\theta})tan{\theta}}{-(1-tan{\theta})})^2(−(1−tanθ)(1−tanθ)tanθ)2

=tan^2{\theta}tan2θ

Hence, LHS=RHS, thus proved.