InterviewSolution
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InterviewSolution
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सिद्ध कीजिए - `tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))=(pi)/(4)-(1)/(2)cos^(-1)x, -(1)/(sqrt2)le x le1.` |
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Answer» `L.H.S.=tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))` `x=cos2theta` रखने पर, `=tan^(-1)[(sqrt(1+cos2theta)-sqrt(1-cos 2theta))/(sqrt(1+cos 2theta)+sqrt(1-cos 2theta))]` `=tan^(-1)[(sqrt(1+(2cos^(2)theta-1))-sqrt(1-(1-2sin^(2)theta)))/(sqrt(1+(2cos^(2)theta-1))+sqrt(1-(1-2sin^(2)theta)))],` `" "[because cos 2theta=2cos^(2)theta-1=1-2sin^(2)theta]` `=tan^(-1)[(sqrt(2cos^(2)theta)-sqrt(2sin^(2)theta))/(sqrt(2cos^(2)theta)+sqrt(2sin^(2)theta))]` `=tan^(-1)[(sqrt2cos theta-sqrt2sin theta)/(sqrt2cos theta+sqrt2sin theta)]` `=tan^(-1)[(cos theta=sin theta)/(cos theta+sin theta)]` `=tan^(-1)[(1-tan theta)/(1+tan theta)],` [अंश और हर को `cos theta` से भाग देने पर ] `=tan^(-1)[(tan.(pi)/(4)-tantheta)/(1+tan.(pi)/(4).tan theta)],` `" "[because tan (A-B)=(tanA-tanB)/(1+tan Atan B)]` `=tan^(-1)[tan((pi)/(4)-theta)]` `=(pi)/(4)-theta` `=(pi)/(4)-(1)/(2)cos^(-1)x` `[because x=cos 2theta rArr 2 theta cos^(-1)x rArr theta=(1)/(2)cos^(-1)x]" "=(x)/(2)` = R.H.S. यही सिद्ध करना था । |
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