Question

দেখাও যে, $tan2A = \frac{2tanA}{1-tan^2A}$ যদি $A=30^\circ$ হয়।

Answer 1

দেওয়া আছে,
$A=30^\circ$

Left Hand Side
$=tan2A$

$=tan\left(2 \times 30^\circ\right)$

$=tan60^\circ$

$=\sqrt3$

Right Hand Side
$= \frac{2tanA}{1-tan^2A}$

$= \frac{2 tan30^\circ}{1-tan^230^\circ}$

$= \frac{2 \times \frac{1}{\sqrt3}}{1-\left(\frac{1}{\sqrt3}\right)^2}$

$= \frac{\frac{2}{\sqrt3}}{1-\frac{1}{3}}$

$= \frac{\frac{2}{\sqrt3}}{\frac{3-1}{3}}$

$= \frac{\frac{2}{\sqrt3}}{\frac{2}{3}}$

$= \frac{2}{\sqrt3} \times \frac{3}{2}$

$= \frac{2}{\sqrt3} \times \frac{\sqrt3 \times \sqrt3}{2}$

$= \sqrt3$

$\therefore$ Left Hand Side = Right Hand Side

অর্থাৎ $tan2A = \frac{2tanA}{1-tan^2A}$