Question #d7b02

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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: $sin^2 theta = 1 -cos^2 theta$

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

$= cos theta /sin theta$

$= cot theta$