Question #3ad93

1 Answer
Feb 7, 2018

Given #tan theta + sin theta=m#, #tan theta-sin theta=n#

Now

#m^2-n^2#

#=(tantheta+sintheta)^2-(tantheta-sintheta)^2#

#=4tanthetasintheta#

#=4sqrt(tan^2thetasin^2theta)#

#=4sqrt(tan^2theta(1-cos^2theta))#

#=4sqrt(tan^2theta-tan^2thetacos^2theta)#

#=4sqrt(tan^2theta-sin^2theta/cos^2thetacos^2theta)#

#=4sqrt(tan^2theta-sin^2theta)#

#=4sqrt((tantheta+sintheta)(tantheta-sin theta))#

#=4sqrt(mn)#