Question

if tan theta+sin theta = m and tan theta-sin theta=n show that m²-n²=4√mn
(question from trignometery chapter)

Answer 1

tanθ+sinθ=m
tanθ-sinθ=n
∴, m+n=tanθ+sinθ+tanθ-sinθ
=2tanθm-n=tanθ+sinθ-tanθ+sinθ
=2sinθmn=(tanθ+sinθ)(tanθ-sinθ)     
=tan²θ-sin²θ∴, m²-n²
=(m+n)(m-n)=2tanθ.2sinθ
=4sinθtanθ4√mn
=4√(tan²θ-sin²θ)
=4√(sin²θ/cos²θ-sin²θ)
=4√sin²θ(1/cos²θ-1)
=4sinθ√(1-cos²θ)/cos²θ
=4sinθ/cosθ√sin²θ [∵, sin²θ+cos²θ=1]=4sinθtanθ∴, LHS=RHS (proved)