please answer the following questions
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Answered by
1
Answer:
Step-by-step explanation:
tanθ+sinθ=m
tanθ-sinθ=n
∴ m+n=tanθ+sinθ+tanθ-sinθ=2tanθ
m-n=tanθ+sinθ-tanθ+sinθ=2sinθ
mn=(tanθ+sinθ)(tanθ-sinθ)
=tan²θ-sin²θ
∴ m²-n²
=(m+n)(m-n)
=2tanθ.2sinθ
=4sinθtanθ
4√mn
=4√(tan²θ-sin²θ)
=4√(sin²θ/cos²θ-sin²θ)
=4√sin²θ(1/cos²θ-1)
=4sinθ√(1-cos²θ)/cos²θ
=4sinθ/cosθ√sin²θ [∵ sin²θ+cos²θ=1]
=4sinθtanθ
= 4√mn
∴ LHS=RHS (proved)
PLEASE MARK AS BRAINLIEST!!!
Answered by
0
Answer:
Step-by-step explanation:
tanθ-sinθ=n
∴ m+n=tanθ+sinθ+tanθ-sinθ=2tanθ
m-n=tanθ+sinθ-tanθ+sinθ=2sinθ
mn=(tanθ+sinθ)(tanθ-sinθ)
=tan²θ-sin²θ
∴ m²-n²
=(m+n)(m-n)
=2tanθ.2sinθ
=4sinθtanθ
4√mn
=4√(tan²θ-sin²θ)
=4√(sin²θ/cos²θ-sin²θ)
=4√sin²θ(1/cos²θ-1)
=4sinθ√(1-cos²θ)/cos²θ
=4sinθ/cosθ√sin²θ [∵ sin²θ+cos²θ=1]
=4sinθtanθ
= 4√mn
∴ LHS=RHS (proved)
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