Question

# Find the value of trigonometric equation ${\text{3cose}}{{\text{c}}^2}\left( {{{60}^0}} \right) - 2{\cot ^2}\left( {{{30}^0}} \right) + {\sec ^2}\left( {{{45}^0}} \right)$.

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Hint: In this question, we use trigonometric values to solve problems. Like ${\text{cosec}}\left( {{{60}^0}} \right) = \dfrac{2}{{\sqrt 3 }},\cot \left( {{{30}^0}} \right) = \sqrt 3$ and $\sec \left( {{{45}^0}} \right) = \sqrt 2$ put these trigonometric values in question and after some calculation we can get the required answer.

Given, ${\text{3cose}}{{\text{c}}^2}\left( {{{60}^0}} \right) - 2{\cot ^2}\left( {{{30}^0}} \right) + {\sec ^2}\left( {{{45}^0}} \right)$
${\text{cosec}}\left( {{{60}^0}} \right) = \dfrac{2}{{\sqrt 3 }},\cot \left( {{{30}^0}} \right) = \sqrt 3$ and $\sec \left( {{{45}^0}} \right) = \sqrt 2$
Put these trigonometric values in ${\text{3cose}}{{\text{c}}^2}\left( {{{60}^0}} \right) - 2{\cot ^2}\left( {{{30}^0}} \right) + {\sec ^2}\left( {{{45}^0}} \right)$
$\Rightarrow 3 \times {\left( {\dfrac{2}{{\sqrt 3 }}} \right)^2} - 2 \times {\left( {\sqrt 3 } \right)^2} + {\left( {\sqrt 2 } \right)^2}$
$\Rightarrow 3 \times \dfrac{4}{3} - 2 \times 3 + 2 \\ \Rightarrow 4 - 6 + 2 \\ \Rightarrow 6 - 6 \\ \Rightarrow 0 \\$
So, the value of ${\text{3cose}}{{\text{c}}^2}\left( {{{60}^0}} \right) - 2{\cot ^2}\left( {{{30}^0}} \right) + {\sec ^2}\left( {{{45}^0}} \right)$ is 0.