Question

Evaluate the following
$\left( {{\operatorname{cosec}}^{2}}{{45}^{o}}{{\sec }^{2}}{{30}^{o}} \right)\left( {{\sin }^{2}}{{30}^{o}}+4{{\cot }^{2}}{{45}^{o}}-{{\sec }^{2}}{{60}^{o}} \right)$

Answer 1

Hint:First of all, consider the expression given in the question. Now make the table for trigonometric ratios of general angles. Now, from that find the values of $\sin {{30}^{o}},\operatorname{cosec}{{45}^{o}},\sec {{60}^{o}},\sec {{30}^{o}}$ and $\cot {{45}^{o}}$ and substitute these in the given expression to get the required answer.

Complete step-by-step answer:
In this question, we have to find the value of the expression
$\left( {{\operatorname{cosec}}^{2}}{{45}^{o}}{{\sec }^{2}}{{30}^{o}} \right)\left( {{\sin }^{2}}{{30}^{o}}+4{{\cot }^{2}}{{45}^{o}}-{{\sec }^{2}}{{60}^{o}} \right)$
Let us consider the expression given in the question.
$E=\left( {{\operatorname{cosec}}^{2}}{{45}^{o}}{{\sec }^{2}}{{30}^{o}} \right)\left( {{\sin }^{2}}{{30}^{o}}+4{{\cot }^{2}}{{45}^{o}}-{{\sec }^{2}}{{60}^{o}} \right)....\left( i \right)$
Now, we have to find the values of $\sin {{30}^{o}},\operatorname{cosec}{{45}^{o}},\sec {{60}^{o}},\sec {{30}^{o}}$ and $\cot {{45}^{o}}$.
Let us make the table for trigonometric ratios of general angles like ${{0}^{o}},{{30}^{o}},{{45}^{o}},{{60}^{o}},{{90}^{o}}$ and find the required values.

From the above table, we get, $\operatorname{cosec}{{45}^{o}}=\sqrt{2}$. By substituting this in equation (i), we get,
$E=\left( {{\left( \sqrt{2} \right)}^{2}}.{{\sec }^{2}}{{30}^{o}} \right)\left( {{\sin }^{2}}{{30}^{o}}+4{{\cot }^{2}}{{45}^{o}}-{{\sec }^{2}}{{60}^{o}} \right)$
Also from the above table, we get $\sec {{30}^{o}}=\dfrac{2}{\sqrt{3}}$. By substituting this in the above equation, we get, $E=\left( {{\left( \sqrt{2} \right)}^{2}}.{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}} \right)\left( {{\sin }^{2}}{{30}^{o}}+4{{\cot }^{2}}{{45}^{o}}-{{\sec }^{2}}{{60}^{o}} \right)$
From the table, we also get, $\sin {{30}^{o}}=\dfrac{1}{2}$. By substituting this in the above equation, we get,
$E=\left( {{\left( \sqrt{2} \right)}^{2}}.{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}} \right)\left( {{\left( \dfrac{1}{2} \right)}^{2}}+4{{\cot }^{2}}{{45}^{o}}-{{\sec }^{2}}{{60}^{o}} \right)$
From the table, we also get, $\cot {{45}^{o}}=1$. By substituting this in the above equation, we get,
$E=\left( {{\left( \sqrt{2} \right)}^{2}}.{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}} \right)\left( {{\left( \dfrac{1}{2} \right)}^{2}}+4{{\left( 1 \right)}^{2}}-{{\sec }^{2}}{{60}^{o}} \right)$
From the table, we also get, $\sec {{60}^{o}}=2$. By substituting this in the above equation, we get,
$E=\left( {{\left( \sqrt{2} \right)}^{2}}.{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}} \right)\left( {{\left( \dfrac{1}{2} \right)}^{2}}+4{{\left( 1 \right)}^{2}}-{{\left( 2 \right)}^{2}} \right)$
By simplifying the above equation, we get,
$E=\left[ 2\left( \dfrac{4}{3} \right) \right]\left[ \dfrac{1}{4}+4-4 \right]$
$E=\left( \dfrac{8}{3} \right)\left( \dfrac{1}{4} \right)$
$E=\dfrac{2}{3}$
Hence, we get the value of the expression $\left( {{\operatorname{cosec}}^{2}}{{45}^{o}}{{\sec }^{2}}{{30}^{o}} \right)\left( {{\sin }^{2}}{{30}^{o}}+4{{\cot }^{2}}{{45}^{o}}-{{\sec }^{2}}{{60}^{o}} \right)$ as $\dfrac{2}{3}$.

Note: In these types of questions, students just need to remember the values of $\sin \theta$ and $\cos \theta$ at various angles like ${{0}^{o}},{{30}^{o}},{{60}^{o}},{{45}^{o}},$ etc. and they can find all other trigonometric ratios using them. For example, they can find $\operatorname{cosec}{{45}^{o}}$ by using $\dfrac{1}{\sin {{45}^{o}}},\sec {{30}^{o}}$ by using $\dfrac{1}{\cos {{30}^{o}}},\cot {{45}^{o}}$ by using $\dfrac{\cos {{45}^{o}}}{\sin {{45}^{o}}}$ and $\sec {{60}^{o}}$ by using $\dfrac{1}{\cos {{60}^{o}}}$.