Question

Prove that $ta{n}^{2}A+co{t}^{2}A+2=cose{c}^{2}Ase{c}^{2}A$

Answer 1

Determine the proof of the expression that is $ta{n}^{2}A+co{t}^{2}A+2=cose{c}^{2}Ase{c}^{2}A$

Use formula:

$$\begin{gathered} {\sec }^2\theta -{\tan }^2\theta =1 \\ {\mathrm{cosec}}^2\theta -{\cot }^2\theta =1 \end{gathered}$$

Solve the L.H.S part:

$$\begin{gathered} ta{n}^{2}A+co{t}^{2}A+2=(1+ta{n}^{2}A)+(1+co{t}^{2}A) \\ \Rightarrow ={\sec }^{2}A+{\mathrm{cosec}}^{2}A \\ \Rightarrow =\frac{1}{{\cos }^{2}A}+\frac{1}{{\sin }^{2}A} \\ \Rightarrow =\frac{{\sin }^{2}A+co{s}^{2}A}{{\cos }^{2}A{\sin }^{2}A}\mathbf{\because }\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{}\mathit{A}\mathbf{}\mathbf{=}1 \\ \Rightarrow =\frac{1}{{\cos }^{2}A{\sin }^{2}A} \\ \Rightarrow =\cos e{c}^{2}Ase{c}^{2}A \end{gathered}$$

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