Question #d7b02

Question

1124 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-08T06:59:25

See proof below

We need

#cos^2x+sin^2x=1#

#cscx=1/sinx#

#cotx=cosx/sinx#

Therefore,

#RHS=cos thetasintheta+cos^3thetacsctheta#

#=costhetasintheta+cos^3theta/sintheta#

#=(costhetasin^2theta+cos^3theta)/sintheta#

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

#=costheta/sintheta#

#=cottheta#

#=LHS#

#Q.E.D#

Answer 2 · 2017-03-08T07:01:26

see explanation

Let we take a right hand side to prove left hand side.

#cos theta sin theta + cos^3 theta csc theta = cos theta sin theta + cos^3 theta * 1/sin theta#

multiply with

#= (cos theta sin theta sin theta+ cos^3 theta) /sin theta#

#= (cos theta sin^2 theta + cos^3 theta) /sin theta#

#= (cos theta (1-cos^2 theta )+ cos^3 theta) /sin theta#

note: #sin^2 theta = 1 -cos^2 theta#

#= (cos theta - cancelcos^3 theta + cancelcos^3 theta) /sin theta#

#= cos theta /sin theta#

# = cot theta#