The expression of \frac{\cot \\theta+\cosec \\theta-1}{\cot \\theta+\cosec \\theta+1} is equal to

Correct Answer is : 1+cosθsinθ

\frac{1+\cos \\theta}{\sin \\theta}

\frac{1-\cos \\theta}{\sin \\theta}

\frac{\cot \\theta+1}{\cosec \\theta}

\frac{\cot \\theta-1}{\cosec \\theta}

SSC CGL Tier II 1 Dec 2016 Paper 1

Question : 95 of 100

Marks: +1, -0

The expression of

{\cot \θ+\cosec \θ-1}/{\cot \θ+\cosec \θ+1}

is equal to

Solution:

(*)

{\cot \θ+\cosec \θ-1}/{\cot \θ-\cosec \θ+1}

(we have taken

({\cot \θ-\cosec \θ+1})

instead of

({\cot \θ+\cosec \θ+1})

as denominator)

= {\cot \θ+\cosec\θ-\ (\cosec ^{2}\θ-\cot ^{2} \θ)}/{\cot\θ-\cosec\θ+1}

= {\cot \θ+\cosec\θ-\ (\cosec \θ+\cot \θ)(\cosec \θ-\cot \θ)}/{\cot\θ-\cosec\θ+1}

= {(\cot \θ+\cosec\θ)-\ (1-\cosec \θ+\cot \θ)}/{\cot\θ-\cosec\θ+1}

= \cot \θ+\cosec\θ

= {\cos \θ}{\sin \θ}+ {1}/{\sin \θ}

= {\cos θ+1}/{\sin θ}

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