Question

If (sec A – tan A) = x then prove that $\frac{1+x^2}{1-x^2}$= cosec A.

Answer 1

Given: $\sec A-\tan A=x.....\left(1\right)$
We know
$$\begin{gathered} {\sec }^{2}A-{\tan }^{2}A=1 \\ \Rightarrow \left(\sec A+\tan A\right)\left(\sec A-\tan A\right)=1\left[a^2-b^2=\left(a-b\right)\left(a+b\right)\right] \\ \Rightarrow \left(\sec A+\tan A\right)x=1\left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow \sec A+\tan A=\frac{1}{x}.....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
$$\begin{gathered} \sec A-\tan A+\sec A+\tan A=x+\frac{1}{x} \\ \Rightarrow 2\sec A=\frac{x^2+1}{x} \\ \Rightarrow \sec A=\frac{x^2+1}{2x}.....\left(3\right) \end{gathered}$$
Subtracting (1) from (2), we get
$$\begin{gathered} \sec A+\tan A-\sec A+\tan A=\frac{1}{x}-x \\ \Rightarrow 2\tan A=\frac{1-x^2}{x} \\ \Rightarrow \tan A=\frac{1-x^2}{2x}.....\left(4\right) \end{gathered}$$
Dividing (3) by (4), we get
$$\begin{gathered} \frac{{\displaystyle \frac{1+x^2}{2x}}}{{\displaystyle \frac{1-x^2}{2x}}}=\frac{\sec A}{\tan A} \\ \Rightarrow \frac{1+x^2}{1-x^2}=\frac{{\displaystyle \frac{1}{\cos A}}}{{\displaystyle \frac{\sin A}{\cos A}}} \\ \Rightarrow \frac{1+x^2}{1-x^2}=\frac{{\displaystyle 1}}{{\displaystyle \sin A}} \\ \Rightarrow \frac{1+x^2}{1-x^2}=\mathrm{cosec}A \end{gathered}$$