Question

Prove the following identity:
tanA/secA+1 + cotA/cosecA+1 = cosecA + secA - secA.tanA

Answer 1

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