Question

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Answer 1

Step-by-step explanation:

Given prove that (cos4x *sin3x - cos2x* sinx) / (sin4x* sinx + cos6x* cosx) = tan 2x

 Consider the given , multiply and divide by 2 we get

=  2(cos4x* sin3x - cos2x* sinx) / 2(sin4x* sinx + cos6x* cosx)

= 2cos4x* sin3x - 2cos2x* sinx) / 2sin4x *sinx + 2cos6x* cosx

 We know that, 2 cos a sin b = sin (a + b) - sin (a - b)

                          2 sin a sin b = cos(a - b) - cos(a + b)

                          2 cos a cos b = cos(a + b) + cos(a - b)

 =  sin (4x + 3x) - sin (4x - 3x) - sin (2x + x) - sin (2x - x)  /  cos(4x - x) - cos(4x + x) + cos(6x + x) + cos(6x - x)

 =   sin 7x - sin x - sin 3x + sin x / cos 3 x - cos 5 x + cos 7x + cos 5x

=   sin 7 x - sin 3 x / cos 3 x + cos 7 x

We Know That , sin a - sin b = 2 sin ($\frac{a+b}{2}$) * cos ($\frac{a-b}{2}$)

                          cos a + cos b = 2 cos($\frac{a+b}{2}$) * cos($\frac{a-b}{2}$)

=  2 cos (7x + 3x / 2)* sin (7x - 3 x /2)  /  2 cos (3x + 7x/2) *cos (7x - 3x / 2)

= 2 cos 5 x* sin 2 x / 2 cos 5 x *cos 2 x

 =  sin 2x / cos 2 x

= tan 2 x

Hence Proved

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