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Question

Which of the following options is equal to sin θ+1cos θcos θ1+sin θ?


A

cosθ1sin θ

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B

1+sinθcosθ

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C

1+sinθ1sinθ

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D

tanθ+secθ

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Solution

The correct option is D

tanθ+secθ


sinθ+1cosθcosθ1+sinθ

Dividing the numerator and denominator by cos θ we get,

sinθcosθ+1cos θcosθcosθcosθcosθ1cosθ+sinθcosθ

=tanθ+secθ11secθ+tanθ

(tanθ+secθ)(sec2θtan2θ)1secθ+tanθ....(1+tan2θ=sec2θ)

=(tanθ+secθ)(secθtanθ)(secθ+tanθ)1secθ+tanθ

=(tanθ+secθ)(1secθ+tanθ)(1secθ+tanθ)

=tanθ+secθ

=sinθcosθ+1cosθ

=sinθ+1cosθ

[Multiplying the numerator and denominator by (1sinθ)]
=1+sinθcosθ×1sinθ1sinθ

=1sin2θcosθ(1sinθ)

=cos2θcosθ(1sinθ)

=cosθ1sinθ


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