If tanA+sinA=m and tanA - sinA =n, show that m^2−n^2= 4√mn

1.	If tanA+sinA=m and tanA - sinA =n, show that m^2−n^2= 4√mn
Answer» Lhs(tan A+sin A )(tan A-sin A) _(tanA+sin A)(tan A-sin A)=(2tan A)(2sin A)=4tan A sin ARhs 4√(tanA+sin A)(tanA-sinA)4√tan Square A-sin square A4√tan square A_sin square A4√sin square A/cos square A_sin square A4√sin square A (1/cos square A-1)4√sin square A (sec square A-1)4√sin square A tan square A4 sin A tan AHence proved

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