Question

Show that $\frac{1-ta{n}^{2}A}{1+ta{n}^{2}A}=2co{s}^{2}A-1$

Answer 1

$L.H.S=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$

$=\frac{1-{\tan }^{2}A}{{\sec }^{2}A}$ $\because (1+{\tan }^{2}A={\sec }^{2}A)$

$={\cos }^{2}A-\frac{{\tan }^{2}A}{{\sec }^{2}A}$

$={\cos }^{2}A-{\sin }^{2}A$

$={\cos }^{2}A-(1-{\cos }^{2}A)$

$=2{\cos }^{2}A-1$

$=R.H.S$