Question

If tanθ + sinθ = m and tanθ - sinθ = n , show that m2 - n2 =4 root mn

Answer 1

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)

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)