Question

Tantheta + sintheta =a and tantheta -sintheta =b then prove it a2 - b2 = root 4ab

Answer 1

a=(tanΘ+sinΘ)
a^2=(tanΘ+sinΘ)^2
      =tan^2Θ+sin^2Θ+2tanΘsinΘ
b=(tanΘ-sinΘ)
b^2=(tanΘ-sinΘ)^2
      =tan^2Θ+sin^2Θ-2tanΘsinΘ
To prove a^2-b^2=√16ab
LHS
a^2-b^2
(tan^2Θ+sin^2Θ+2tanΘsinΘ)-(tan^2Θ+sin^2Θ-2tanΘsinΘ)
tan^2Θ+sin^2Θ+2tanΘsinΘ-tan^2Θ-sin^2Θ+2tanΘsinΘ
4tanΘsinΘ
RHS
√16ab
√16(tanΘ+sinΘ)(tanΘ-sinΘ)
√16(tan^2Θ-sin^2Θ)
√16[(sin^2Θ/cos^2)-sin^2Θ]
√16[sin^2Θ(1-cos^2Θ)]/cos^2Θ              {sin^2Θ+cos^2Θ=1}
√16[sin^2Θ·(sin^2Θ/cos^2Θ)]
√16tan^2Θsin^2Θ
4tanΘsinΘ