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Question
`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`
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Solution
LHS =` (1+ cos theta + sin theta)/(1+ cos theta-sin theta)`
=` ({(1+cos theta)+ sin theta}{(1+ cos theta)+ sin theta})/({(1+ cos theta )-sin theta}{(1+ cos theta )+ sin theta}) {"Multiplying the numerator and denominator by "(1 + costheta +sin theta}`
=`({(1+ cos theta)+ sin theta}^2)/({(1+ cos theta )^2-sin ^2 theta})`
=`(1+ cos^2 theta + 2 cos theta + sin ^2 theta + 2 sin theta (1+ cos theta))/(1+ cos^2 theta + 2 cos theta - sin ^2 theta)`
=`(2+2 cos theta + 2 sin theta (1+ cos theta))/(1+ cos ^2 theta + 2 cos theta -(1-cos^2 theta))`
=`(2(1+ cos theta)+2sin theta (1+ cos theta))/(2 cos^2 theta+2 cos theta)`
=`(2(1+ cos theta) (1+ sin theta))/( 2 cos theta (1+ cos theta))`
=`(1+sin theta)/cos theta`
= RHS
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