Question

sin thita+1-cos thita/cos thita-1+sin thita=1+sin thita/cos thita
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Answer 1

Answer:

Step-by-step explanation:

Let theta = A

L.H.S :

= (sin A + 1 - cos A)/(cos A - 1 + sin A)

= {(sin A + 1 - cos A)(cos A + 1 - sin A)}/{(cos A - 1 + sin A)(cos A + 1 - sin A)}

= (sin A cos A - sin^2 A  + sin A - cos^2 A  + sin A cos A - cos A + cos A - sin A + 1)/(cos^2 A - sin A cos A + cos A + sin A cos A - sin^2 A + sin A - cos A + sin A -1)

= (2 sin A cos A)/(2 sin A - 2 sin^2 A )

= (2 sin A cos A )/(2 sin A(1-sin A))

= cos A / (1 - sin A)

= cos A(1 + sin A)/{(1 - sin A)(1 + sin A)}

= cos A(1 + sin A)/1 - 2 sin^2 A

= cos A(1 + sin A)/cos^2 A

=(1 + sin A)/cos A == R.H.S.

Answer 2

LHS = sin thita.cos thita - sin square theta + sin theta - cos square theta + sin theta . cos theta - cos theta+ cos theta - sin theta +1 / cos square theta - sin theta . cos theta + cos theta + sin thita .cos theta - sin square theta + sin theta - cos theta + sin thita -1

= (2 sin thita .cos thita)/2 sin thita -2 sin square theta

= (2 sin thita .cos theta)/2 sin thita (1- sin thita)

= cos theta /1- sin thita

= cos theta (1+ sin thita)/(1- sin thita)(1+ sin thita)

= cos theta (1+ sin thita)/1-2 sin square theta

= cos theta (1+ sin thita )/ cos square theta

= 1+ sin thita / cos theta = RHS

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