Question

Show that : $\cot (A+{15}^{\circ })-\tan (A-{15}^{\circ })=\frac{4\cos 2A}{1+2\sin 2A}$

Answer 1

$cot(A+{15}^{\circ })-tan(A-{15}^{\circ })=\frac{4cos2}{1+2sin2A}$

L.H.S cot(A+15) - tan(A-15)

$\Rightarrow \frac{cos(A+15)}{sin(A+15)}-\frac{sin(A-15)}{cos(A-15)}$

$\Rightarrow \frac{cos(A-15)cos(A+15)-sin(A-15)sin(A+15)}{sin(A+15)cos(A-15)}$

$\Rightarrow \frac{2cos2\left(A\right)}{2sin(A+15)\left(cos\right(A-15\left)\right)}$

$\Rightarrow \frac{2cos2A}{sin2A+sin30}$

$\Rightarrow \frac{2cos2A}{[\frac{1}{2}+sin2A]}$

$\Rightarrow \frac{2cos2A.2}{[1+2sin2A]}=\frac{4.cos2A}{1+2sin2A}$ R.H.S