Question

Pls solve these ques

Answer 1

Answer:

(1+cos4x)/(cotx-tanx)=1/2sin4x

LHS

=(1+cos4x)/(cotx-tanx)

now, in numerator using identity cos2x=2cos^2x-1

then we get (1+cos2(2x))/(cotx-tanx)=(1+2cos^2(2x)-1)/(cotx-tanx)=((2cos^2(2x))/(cotx-tanx)

now in denominator using the identity tanx=sinx/cosx,and cotx=cosx/sinx we get

(2os2x)/(cosx/sinx-sinx/cosx)=((2cos^2(2x))/(cos^2x-sin^2x)/sinx.cosx

now again by using cos^2x-sin^2x=cos2x

we get( (2cos^2(2x))/(cos2x)/(sinx.cosx)after solving further we get 2cos^2(2x)×sinx.cosx/cos2x = (2sinx.cosx).(cos2x)

now because 2sinx.cosx=sin2x

so we have (2sinx.cosx).(cos2x)=sin2x.cos2x = 1/2×2sin2x.cos2x = 1/2.sin4x=r.hs ,proved.