Question

Prove the identity:
cos^​2theta + tan²theta - 1/sin^​2theta = tan²theta

Answer 1

$$\begin{gathered} \mathrm{To}\mathrm{Prove}:- \\ \frac{{\cos }^2\mathrm{A}+{\tan }^2\mathrm{A}-1}{{\sin }^2\mathrm{A}}={\tan }^2\mathrm{A} \\ \mathrm{LHS} \\ \frac{{\cos }^2\mathrm{A}+{\tan }^2\mathrm{A}-1}{{\sin }^2\mathrm{A}}=\frac{-{\sin }^2\mathrm{A}+{\displaystyle \frac{{\sin }^2\mathrm{A}}{{\cos }^2\mathrm{A}}}}{{\sin }^2\mathrm{A}}\left(\mathrm{Since}{\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}=1\right) \\ =\frac{{\sin }^2\mathrm{A}\left({\displaystyle \frac{1}{{\cos }^2\mathrm{A}}-1}\right)}{{\sin }^2\mathrm{A}}=\frac{1-{\cos }^2\mathrm{A}}{{\cos }^2\mathrm{A}} \\ =\frac{{\sin }^2\mathrm{A}}{{\cos }^2\mathrm{A}}={\tan }^2\mathrm{A}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{Proved}. \end{gathered}$$