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This question belongs to Arithmetic Ability Trigonometry
Trigonometry
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If $${\text{sin}}\left( {{{90}^ \circ } - \theta } \right)$$   + $${\text{cos}}\theta $$  = $$\sqrt 2 {\text{cos}}\left( {{{90}^ \circ } - \theta } \right){\text{,}}$$    then the value of $${\text{cosec}}\theta $$   is?

Answer & Solution
Correct Answer: Option C
$$\eqalign{ & {\text{sin}}\left( {{{90}^ \circ } - \theta } \right) + {\text{cos}}\theta = \sqrt 2 {\text{cos}}\left( {{{90}^ \circ } - \theta } \right) \cr & \Rightarrow {\text{cos}}\theta + {\text{cos}}\theta = \sqrt 2 \sin \theta \cr & \Rightarrow \frac{{2\cos \theta }}{{\sin \theta }} = \sqrt 2 \cr & \Rightarrow \cot \theta = \frac{{1 \to {\text{B}}}}{{\sqrt 2 \to {\text{P}}}} \cr & {\text{So, H}} \to \text{alignment} \cr & \therefore {\text{cosec}}\theta = \frac{{\text{H}}}{{\text{P}}} = \frac{{\sqrt 3 }}{{\sqrt 2 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{3}{2}} \cr} $$
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