Question

if m=tanθ+sinθ, tanθ-sinθ=n and m ≠n then prove that m²-n²=4√mn

Answer 1

tanA+sinA=m and tanA-sinA
m squre-n squre =4√mn
taking LHS
m squre-n squre
(m+n) (m-n)
(tanA+sinA+tanA-sinA )(tanA+sinA-tan+sinA)
(2tanA)(2sinA)
4tanAsinA
taking RHS
4√mn
4√(tanA+sinA)(tanA-sinA)
4√tan squreA-sin squreA
4√sin squreA\cos squreA-sin squreA
4√(sin squreA- cos squreA- sin squreA)\cos squreA
4√sin squre(1/cos squreA-1)
4√sin squreA×tan squreA
4sinA tanA
LHS=RHS

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)