Question

Please solve this problem​

Answer 1

Answer:

Given:

tangentθ+sineθ=m

tangentθ-sineθ=n

∴ m+n=tanθ+sinθ+tanθ-sinθ

=2tanθ

∴m-n=tanθ+sinθ-tanθ+sinθ

=2sinθ

and, mn=(tanθ+sinθ)(tanθ-sinθ)

     =tan²θ-sin²θ

∴ mn=m²-n²

Thus,

LHS=m²–n²=(m+n)(m-n)

=2tanθ.2sinθ

=4sinθtanθ

RHS=4√mn=4√(tan²θ-sin²θ)

=4√(sin²θ/cos²θ-sin²θ)

[∵ tanθ=sinθ/cosθ]

=4√sin²θ(1/cos²θ-1)

=4sinθ√(1-cos²θ)/cos²θ

=4sinθ/cosθ√sin²θ [∵ sin²θ+cos²θ=1]

∴ LHS=4sinθtanθ=RHS

∵LHS=RHS

(Hence, proved)

Answer 2

Step-by-step explanation:

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