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Question

Evaluate: cosθsinθ+1cosθ+sinθ1.

A
cosecθ+cotθ.
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B
cosθ+cotθ.
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C
cosecθ+tanθ.
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D
sinθ+cotθ.
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Solution

The correct option is D cosecθ+cotθ.
Given that cosθsinθ+1cosθ+sinθ1

divide both Numerator and Denominator by sinθ

=cosθsinθ+1sinθcosθ+sinθ1sinθ

=cotθ+coescθ1cotθ+1coescθ

=cotθ+coescθ(coesc2θcot2θ)cotθ+1coescθ

=cotθ+coescθ(1(coescθcotθ)cotθcoescθ+1

=coescθ+cotθ.

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