(\sec \theta+\tan \theta)^{2}+\frac{1+cosec \theta}{1-cosec \theta} 0^{\circ} < \theta < 90^{\circ} is 0^{\circ} < \theta < 90^{\circ} is

SSC CGL 18 Aug 2021 Shift 3 Paper

Question : 74 of 100

Marks: +1, -0

(\sec θ+\tan θ)^{2}+\;{1+cosec θ}/{1-cosec θ}

0^{∘} < θ < 90^{∘}

0^{∘} < θ < 90^{∘}

Solution:

Given,

(\sec θ+\tan θ)^{2}+\;{(1+cosec θ)}/{(1-cosec θ)}

=(\;{1}/{\cos θ}+\;{\sin θ}/{\cos θ})^{2}+\;{(1+cosec θ)}/{(1-cosec θ)}

\;\; =\;{(1+\sin θ)^{2}}/{\cos ^{2} θ}+\;{(1+cosec θ)}/{(1-cosec θ)}

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

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

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

\;\;{cosec θ+1}/{cosec θ}

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

= \;\; \;{(1+cosec θ)}/{(cosec θ-1)}-\;{(1+cosec θ)}/{(cosec θ-1)}=0

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