Question #2ba1b

1 Answer
Feb 23, 2018

#sin2theta=(2tantheta)/(1+tan^2theta#

#=(2sec2theta)/(1+sec^2 2theta)# #color(red)(" Given "tantheta =cos2theta)#

#=(2sec2thetacos^2 2theta)/(cos^2 2theta+cos^2 2thetasec^2 2theta#

#=(2cos 2theta)/(cos^2 2theta+1#

#=((2(1-tan^2theta))/(1-tan^2theta))/(((1-tan^2 theta)/(1+tan^2theta))^2+1)# # color(red)(" Inserting "cos2theta=(1-tan^2 theta)/(1+tan^2theta)#

#=((2(1-tan^2theta))/(1-tan^2theta))/(((1-tan^2 theta)^2+(1+tan^2theta)^2)/(1+tan^2theta)^2)#

#=(2(1-tan^2theta))/cancel(1-tan^2theta)xx(1+tan^2theta)^cancel2/(((1-tan^2 theta)^2+(1+tan^2theta)^2)#

#=(2(1-tan^2theta)(1+tan^2theta))/((2(1+tan^4theta))#

#=(1-tan^4theta)/(1+tan^4theta)=RHS#