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If m=cos theta-sin theta and n=costheta+...

If `m=cos theta-sin theta` and `n=costheta+sin theta` then show that `sqrt(m/n)+sqrt(n/m)=2/sqrt(1-tan^2 theta)`

Text Solution

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`LHS=((m+n))/(sqrt(mn)).`
Now, `(m+n)=2costheta and nm=(cos^(2)theta-sin^(2)theta)=cos^(2)theta(1-tan^(2)theta).`
` therefore sqrt(mn)=costheta sqrt(1-tan^(2)theta).`
Hence, `LHS =((m+n))/(sqrt(mn))=(2 costheta)/(costheta sqrt(1-tan^(2)theta))=(2)/(sqrt(1-tan^(2)theta))=RHS.`
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