Question

if TanA+sinA=m and tanA-sinA=n,show that m²-n²=4√mn

Answer 1

1: squaring: tan^2a+sin^2a+2*tana*sina= m^2
2: squaring:
Tan^2a+sin^2a-2* tana*sina=n^2
On subtracting
4*tana*sina=m^2-n^2
Eqn 1& 2 multiply
(Tana+sina)(tana-sina)=m*n
Tan^2a-sin^2a=mn
Sin^2a /cos^2a -sin^2a =mn
Sin^2a (1/cos^2a -1)=mn
Sin^2a (sec^2a-1)=mn
Sin^2a*tan^2a=mn
So sina*tana=√mn
So tana*sina =m^2-n^2
m^2-n^2 =4√mn

Answer 2

Answer:

Step-by-step explanation:

tanθ+sinθ=m and tanθ-sinθ=n

∴, m²-n²

=(m+n)(m-n)

=(tanθ+sinθ+tanθ-sinθ)(tanθ+sinθ-tanθ+sinθ)

=(2tanθ)(2sinθ)

=4tanθsinθ

4√mn

=4√(tanθ+sinθ)(tanθ-sinθ)

=4√(tan²θ-sin²θ)

=4√{(sin²θ/cos²θ)-sin²θ}

=4√sin²θ{(1/cos²θ)-1}

=4sinθ√{(1-cos²θ)/cos²θ}

=4sinθ√(sin²θ/cos²θ)

=4sinθ√tan²θ

=4sinθtanθ

∴, LHS=RHS (Proved)