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Questions  

If 0<θ<π2 then  tanθ+secθ1tanθsecθ+1 is equal to 

a
1+sin⁡θ+cos⁡θ
b
1+sin⁡θcos⁡θ
c
1−cos⁡θsin⁡θ
d
tan⁡θ−sec⁡θ

detailed solution

Correct option is B

tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1= tan⁡θ+sec⁡θ−sec2θ-tan2θtan⁡θ−sec⁡θ+1=(sec⁡θ+tan⁡θ)(1−sec⁡θ+tan⁡θ)tan⁡θ−sec⁡θ+1=1+sin⁡θcos⁡θ

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