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InterviewSolution
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If `cosectheta + cottheta=p`, then prove that the `cos theta=(p^2-1)/(p^2+1)` |
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Answer» Given, `cosectheta + cottheta=p` `rArr 1/(sintheta)+ (costheta)/(sintheta)=p` `[therefore cosectheta=1/(sintheta)` and `cottheta=(costheta)/(sintheta)]` `rArr (1+costheta)/(sintheta)=p/1` `rArr (1+costheta)^(2)/(sin^(2)theta)=p^(2)/1` [take square on both sides] `rArr (1+ cos^(2)theta + 2 cos theta)/(sin^(2)theta)=p^(2)/1` Using componendo and dividendo rule, we get `((1+cos^(theta)+2costheta)-sin^(2)theta)/((1+cos^(2)theta+2costheta)+sin^(2)theta)= (p^(2)-1)/(p^(2)+1)` `rArr (1+cos^(2)theta + 2costheta-(1-cos^(2)theta))/(1+2costheta+(cos^(2)theta+sin^(2)theta))=(p^(2)-1)/(p^(2)+1)` `[therefore sin^(2)theta + cos^(2)theta=1]` `rArr (2cos^(2)theta + 2costheta)/(2+2costheta) = (p^(2)-1)/(p^(2)+1)` `rArr (2costheta(costheta+1))/(2(costheta+1))=(p^(2)-1)/(p^(2)+1)` `therefore costheta=(p^(2)-1)/(p^(2)+1)` Hence proved. |
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