InterviewSolution
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InterviewSolution
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Prove that `(i) (1- sin theta)/(1+ sin theta)=(sec theta - tan theta )^(2) ` `(ii) ((1+cos theta))/((1- cos theta))= ("cosec" theta + cot theta)^(2)` |
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Answer» We have `(i) RHS = (sec theta - tan theta)^(2) ` ` =((1)/(cos theta)-(sin theta)/(cos theta))^(2)=((1-sin theta)/(cos theta))^(2)=((1- sin theta)^(2))/(cos^(2)theta)` ` = ((1-sin theta)^(2))/((1-sin^(2)theta))=((1- sin theta)(1-sin theta))/((1+ sin theta)(1-sin theta))` `=((1-sin theta))/((1+sin theta))= LHS. ` ` therefore RHS = LHS. ` `(ii) RHS= ("cosec" theta + cot theta)^(2) =((1)/(sin theta)+(cos theta)/(sin theta))^(2) ` ` =((1+ cos theta)/(sin theta))^(2) = ((1+ cos theta)^(2))/(sin^(2)theta) =((1+ cos theta)(1+ cos theta))/(1- cos^(2)theta)` ` =((1+ cos theta)(1+ cos theta))/((1+ cos theta)(1- cos theta))=((1+ cos theta))/((1- cos theta))= LHS. ` `therefore LHS = RHS. ` |
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