Question

Prove the given identity by using the identity $1+{\cos }^2\theta =\cos e{c}^2\theta$:

${\left(\mathrm{sinA}+\mathrm{secA}\right)}^2+{(\mathrm{cosA}+\mathrm{cosecA})}^2={(1+\mathrm{secA}\mathrm{cosecA})}^2$.

Answer 1

Verify L.H.S and R.H.S.

Let $\mathrm{L}.\mathrm{H}.\mathrm{S}.={(\mathrm{sinA}+\mathrm{secA})}^2+{(\mathrm{cosA}+\mathrm{cosecA})}^2$

$$\begin{gathered} =\left(\frac{1}{\mathrm{cosecA}}+\mathrm{secA}\right)+{\left(\frac{1}{\mathrm{secA}}+\mathrm{cosecA}\right)}^2 \\ ={\left(\frac{1+\mathrm{secA}\mathrm{cosecA}}{\mathrm{cosecA}}\right)}^2+\left(\frac{1+\mathrm{secA}\mathrm{cosecA}}{\mathrm{secA}}\right) \end{gathered}$$

$$\begin{gathered} ={(1+\mathrm{secA}.\mathrm{cosecA})}^2(\frac{1}{{\mathrm{cosec}}^2\mathrm{A}}+\frac{1}{{\sec }^2\mathrm{A}}) \\ ={(1+\mathrm{secA}.\mathrm{cosecA})}^2({\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}) \\ ={(1+\mathrm{secA}.\mathrm{cosecA})}^2\left[\therefore {\sin }^2\mathrm{A}+{\cos }^2\mathrm{B}=1\right] \\ =\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Hence, $L.H.S.=R.H.S.$