Question

If $\sec A=x+\frac{1}{4x}$ prove that $\sec A+\tan A=2x\text{or}\frac{1}{2x}$

Answer 1

$\sec A=x+\frac{1}{4x}$

$\Rightarrow {\sec }^{2}A=x^2+\frac{1}{16x^2}+2.x.\frac{1}{4x}$

$\Rightarrow 1+{\tan }^{2}A=x^2+\frac{1}{16x^2}+\frac{1}{2}$

$\Rightarrow {\tan }^{2}A=x^2+\frac{1}{16x^2}-\frac{1}{2}$

$={(x-\frac{1}{4x})}^2$

$\tan A=\pm (x-\frac{1}{4x})$

$\sec A+\tan A=(x+\frac{1}{4x})\pm (x-\frac{1}{4x})$

$=2x,\frac{1}{2x}$