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Question

Prove that

sec θ1sec θ+1+sec θ+1sec θ1=2 cosec θ


A

=5 cosec θ

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B

=8 cosec θ

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C

=2 cosec θ

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D

=3 cosec θ

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Solution

The correct option is C

=2 cosec θ


sec θ1sec θ+1+sec θ+1sec θ1

=(sec θ1)2+(sec θ+1)2sec2 θ12

=sec θ1+sec θ+1sec2 θ1

=2 sec θtan2 θ

=2 sec θtan θ

=21cos θsin θcos θ

=2sin θ

=2 cosec θ


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