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

Answer

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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.

Complete step-by-step answer:

Given, ${\text{3cose}}{{\text{c}}^2}\left( {{{60}^0}} \right) - 2{\cot ^2}\left( {{{30}^0}} \right) + {\sec ^2}\left( {{{45}^0}} \right)$

As we know the some standard values of trigonometric angles,

${\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 ${\text{3cose}}{{\text{c}}^2}\left( {{{60}^0}} \right) - 2{\cot ^2}\left( {{{30}^0}} \right) + {\sec ^2}\left( {{{45}^0}} \right)$

\[ \Rightarrow 3 \times {\left( {\dfrac{2}{{\sqrt 3 }}} \right)^2} - 2 \times {\left( {\sqrt 3 } \right)^2} + {\left( {\sqrt 2 } \right)^2}\]

Square the following values,

\[

\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.

Note: Whenever we face such types of problems we use some important points. First we write the all values of trigonometric angles should be used in question, then put all trigonometric values in question. So, after some calculation we will get the required answer.

Complete step-by-step answer:

Given, ${\text{3cose}}{{\text{c}}^2}\left( {{{60}^0}} \right) - 2{\cot ^2}\left( {{{30}^0}} \right) + {\sec ^2}\left( {{{45}^0}} \right)$

As we know the some standard values of trigonometric angles,

${\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 ${\text{3cose}}{{\text{c}}^2}\left( {{{60}^0}} \right) - 2{\cot ^2}\left( {{{30}^0}} \right) + {\sec ^2}\left( {{{45}^0}} \right)$

\[ \Rightarrow 3 \times {\left( {\dfrac{2}{{\sqrt 3 }}} \right)^2} - 2 \times {\left( {\sqrt 3 } \right)^2} + {\left( {\sqrt 2 } \right)^2}\]

Square the following values,

\[

\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.

Note: Whenever we face such types of problems we use some important points. First we write the all values of trigonometric angles should be used in question, then put all trigonometric values in question. So, after some calculation we will get the required answer.

Last updated date: 18th Sep 2023

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