Question #ac0e3

Question

895 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-22T16:30:34

see explanation

let we take LHS to prove RHS.

#tan 2theta = sin(2theta) / cos (2theta) #

#sin(2theta) = 2sin theta cos theta, and cos (2theta)= cos^2 theta - sin^2 theta #

therefore,
#tan 2theta= (2sin theta cos theta)/ (cos^2 theta - sin^2 theta)#

divide by #cos^2 theta#

#= (2sin theta cos theta)/cos^2 theta/ ((cos^2 theta - sin^2 theta))/cos^2 theta#

#= (2sin theta cos theta)/cos ^2theta/ (cos^2 theta/cos^2 theta - sin^2 theta/cos^2 theta)#

#= (2sin theta )/cos theta/ (1 - sin^2 theta/cos^2 theta)#

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

Answer 2 · 2017-03-22T16:37:25

#LHS=tan2theta #

#=sin(2theta)/cos(2theta)#

#=(2sinthetacostheta)/(cos^2theta-sin^2theta)#

#=((2sinthetacostheta)/cos^2theta)/((cos^2theta-sin^2theta)/cos^2theta)#

#=((2sintheta)/costheta)/(cos^2theta/cos^2theta-sin^2theta/cos^2theta)#

#= (2tantheta )/(1-tan^2theta)=RHS#