Question

if tanA+sinA=m, and tanA-sinA=n, show that m2-n 2 = 4 under root 'mn'

Answer 1

LHS
(tanA+sinA)²-(tanA-sinA)²
4tanAsinA. :- (a+b)² - (a-b)²= 4ab

now prove RHS
4√(tanA+sinA)(tanA-sinA)
=4√(tan²A-sin²A) √[sin²A/cos²A-sin²A]
=4√sin²A-sin²A cos²A÷cosA=4
sinA/cosA. √1-cos²A
=4tanA. √sin²A cos²A÷cosA=4•sinA/cosA
1-cos²A. ✓sin²A= 4tanA sinA.
(m²-n²)=4√mn thus LHS=RHS proved

Answer 2

here is your answer