Question

prove that sin theta-cos theta+1/sintheta+costheta-1 = costheta/1-sin theta

Answer 1

Answer:

LHS `=(sin theta - cos theta +1)/(sin theta + cos theta -1) ` <br> ` = ((sin theta)/(cos theta)-1 + (1)/(cos theta ))/((sin theta )/(cos theta) +1 - (1)/(cos theta)) ` <br> [ on dividing num. and denom. by ` cos theta ` ] <br> ` =(tan theta -1+ sec theta )/(tan theta +1 - sec theta )=((sec theta + tan theta -1))/((tan theta - sec theta +1)) ` <br> ` = ((sec theta + tan theta )-(sec^(2) theta - tan^(2) theta ))/((tan theta - sec theta +1)) " " [ because 1= sec^(2) theta - tan^(2) theta ] ` <br> ` =((sec theta + tan theta )[1- (sec theta - tan theta )])/((tan theta - sec theta +1)) ` <br> ` = ((sec theta + tan theta)(tan theta - sec theta +1))/((tan theta - sec theta +1))=(sec theta + tan theta). ` <br> ` RHS= (1)/((sec theta - tan theta)) ` <br> ` =(1)/((sec theta - tan theta))xx ((sec theta + tan theta ))/((sec theta + tan theta )) = ((sec theta + tan theta ))/((sec^(2)theta - tan^(2)theta))` <br> `= (sec theta + tan theta) " "[ because sec^(2)theta - tan^(2)theta =1].` <br> Hence,` LHS = RHS.

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Answer 2

Answer:

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