If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m^2 – n^2 = 4sinθ tanθ
Answers
Answered by
0
Namaste... Prove is really easy... First try it again then see the the solution i gave...
Attachments:
Answered by
0
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)
Similar questions