Question

prove that sec a + cosec a = sec a ? cosec a.

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sec \mathrm{A}+\mathrm{cosec}\mathrm{A} \\ =\frac{1}{\cos \mathrm{A}}+\frac{1}{\sin \mathrm{A}} \\ =\frac{\sin \mathrm{A}+\cos \mathrm{A}}{\sin \mathrm{A}.\cos \mathrm{A}} \\ \mathrm{RHS}=\sec \mathrm{A}.\mathrm{cosec}\mathrm{A} \\ =\frac{1}{\cos \mathrm{A}}\times \frac{1}{\sin \mathrm{A}} \\ =\frac{1}{\sin \mathrm{A}.\cos \mathrm{A}} \\ \mathrm{So},\mathrm{LHS}\ne \mathrm{RHS} \\ \mathrm{I}\mathrm{assume}\mathrm{your}\mathrm{question}\mathrm{as}, \\ \mathrm{Prove}{\sec }^2\mathrm{A}+{\mathrm{cosec}}^2\mathrm{A}={\sec }^2\mathrm{A}\times {\mathrm{cosec}}^2\mathrm{A} \\ \mathrm{LHS}={\sec }^2\mathrm{A}+{\mathrm{cosec}}^2\mathrm{A} \\ =\frac{1}{{\cos }^2\mathrm{A}}+\frac{1}{{\sin }^2\mathrm{A}} \\ =\frac{{\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}}{{\cos }^2\mathrm{A}.{\sin }^2\mathrm{A}} \\ =\frac{1}{{\cos }^2\mathrm{A}.{\sin }^2\mathrm{A}} \\ =\frac{1}{{\cos }^2\mathrm{A}}\times \frac{1}{{\sin }^2\mathrm{A}} \\ ={\sec }^2\mathrm{A}.{\mathrm{cosec}}^2\mathrm{A} \\ =\mathrm{RHS} \end{gathered}$$