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If $$\frac{{{{\cos }^2}\theta }}{{{{\cot }^2}\theta - {{\cos }^2}\theta }} = 3,$$ 0° < θ < 90°, then the value of cotθ + cosecθ is:
Answer & Solution
Correct Answer:
Option
A
$$\eqalign{
& \frac{{{{\cos }^2}\theta }}{{{{\cot }^2}\theta - {{\cos }^2}\theta }} = 3 \cr
& \frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta \left( {\frac{{1 - {{\sin }^2}\theta }}{{{{\sin }^2}\theta }}} \right)}} = 3 \cr
& \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = 3 \cr
& \tan \theta = \sqrt 3 \cr
& \theta = {60^ \circ } \cr
& \cot \theta + {\text{cosec}}\,\theta \cr
& = \cot {60^ \circ } + {\text{cosec}}\,{60^ \circ } \cr
& = \frac{1}{{\sqrt 3 }} + \frac{2}{{\sqrt 3 }} \cr
& = \frac{3}{{\sqrt 3 }} \cr
& = \sqrt 3 \cr} $$
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