InterviewSolution
Saved Bookmarks
| 1. |
Prove that `(sin theta - cos theta +1)/(sin theta + cos theta -1) = (1)/((sec theta - tan theta)). ` |
|
Answer» LHS `=(sin theta - cos theta +1)/(sin theta + cos theta -1) ` ` = ((sin theta)/(cos theta)-1 + (1)/(cos theta ))/((sin theta )/(cos theta) +1 - (1)/(cos theta)) ` [ on dividing num. and denom. by ` cos theta ` ] ` =(tan theta -1+ sec theta )/(tan theta +1 - sec theta )=((sec theta + tan theta -1))/((tan theta - sec theta +1)) ` ` = ((sec theta + tan theta )-(sec^(2) theta - tan^(2) theta ))/((tan theta - sec theta +1)) " " [ because 1= sec^(2) theta - tan^(2) theta ] ` ` =((sec theta + tan theta )[1- (sec theta - tan theta )])/((tan theta - sec theta +1)) ` ` = ((sec theta + tan theta)(tan theta - sec theta +1))/((tan theta - sec theta +1))=(sec theta + tan theta). ` ` RHS= (1)/((sec theta - tan theta)) ` ` =(1)/((sec theta - tan theta))xx ((sec theta + tan theta ))/((sec theta + tan theta )) = ((sec theta + tan theta ))/((sec^(2)theta - tan^(2)theta))` `= (sec theta + tan theta) " "[ because sec^(2)theta - tan^(2)theta =1].` Hence,` LHS = RHS. ` |
|