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Question
If `( tan theta + sin theta ) = m and ( tan theta - sin theta ) = n " prove that "(m^2-n^2)^2 = 16 mn .`
Solution
We have `(tan theta + sin theta ) = m and ( tan theta - sin theta )=n`
Now ,LHS = `(m^2-n^2)^2`
=`[(tan^2 theta + sin theta )^2 - "( tan theta - sin theta )^2]^2`
=`[(tan^2 theta + sin^2 theta + 2 tan theta sin theta )-( tan^2 theta + sin^2 theta -2 tan theta sin theta )]^2`
=`[(tan^2 theta +sin^2 theta + 2 tan theta sin theta - tan^2 theta - sin^2 theta+ 2 tan theta sin theta )]^2`
=`(4 tan theta sin theta )^2`
=`16 tan^2 theta sin^2 theta`
=`16 (sin ^2 theta )/(cos^2 theta ) sin^2 theta`
=`16 ((1- cos^2 theta) sin ^2 theta)/ cos^2 theta`
=` 16 [ tan^2 theta (1- cos^2 theta)]`
=`16 (tan^2 theta - tan^2 theta cos^2 theta)`
=`16 (tan^2 theta -(sin^2 theta)/(cos^2 theta) xx cos^2 theta )s`
=`16 ( tan^2 theta - sin^2 theta )`
=`16 (tan theta + sin theta ) ( tan theta - sin theta)`
=`16 mn [(tan theta + sin^theta )( tan theta - sin theta ) =mn]`
=`∴ (m^2 - n^2 )(m^2 - n^2 )^2 = 16 mn`
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