Question

prove :
(Sin 5A COS 2A - sin 6A COS A)/
(sin A sin 2 A- cos 2 A Cos 3 A​)
= tan A

Answer 1

Answer:

$\frac{sin 5A cos 2A -sin 6A cos A}{sin A sin 2A-cos 2A cos 3A}$ = tan A

Step-by-step explanation:

LHS: $\frac{sin 5A cos 2A -sin 6A cos A}{sin A sin 2A-cos 2A cos 3A}$

Multiplying '2' to the numerator and denominator:

$\frac{2(sin 5A cos 2A -sin 6A cos A)}{2(sin A sin 2A-cos 2A cos 3A)}$

$\frac{2sin 5A cos 2A -2sin 6A cos A}{2sin A sin 2A- 2cos 2A cos 3A}$            

$=\frac{sin 7A + sin 3A - sin7A - sin 5A}{cos A - cos 3A -cos 5A - cos A}$  $\left[\begin{array}{ccc}2sin A cos B = sin(A+B) + sin(A-B)\\{2sin A sin B =cos(A-B) -cos(A+B)}\\{2cos A cos B =cos(A+B) +cos(A-B)}\end{array}\right]$ $=\frac{sin 3A - sin 5A}{- cos 3A -cos 5A }$

Multiplying -ve sign to the numerator and denominator:

$=\frac{sin 5A - sin 3A}{ cos 3A + cos 5A }$                            $\left[\begin{array}{ccc}sin A - sin B =2cos(\frac{A+B}{2}) sin(\frac{A-B}{2} ) \\cos A + cos B =2 cos(\frac{A+B}{2} )cos (\frac{A-B}{2} )\\\end{array}\right]$

=  $\frac{2cos(\frac{5A+3A}{2}) sin(\frac{5A - 3A}{2}) } {2cos(\frac{5A+3A}{2} ) cos(\frac{5A-3A}{2}) }$              

= $\frac{2cos( 4A)sin(A) }{2cos(4A)cos(A)}$

= tan A

=RHS

∴ Hence proved