Question

What is the value of $\left( {{{\tan }^4}{{60}^ \circ } - {{\sin }^4}{{90}^ \circ }} \right) - 2{\left( {{{\tan }^2}{{45}^ \circ } - 3\cos {0^ \circ }} \right)^2}$

Answer 1

Hint:- In this question first we need to find the values of different trigonometric ratios involved into$\left( {{{\tan }^4}{{60}^ \circ } - {{\sin }^4}{{90}^ \circ }} \right) - 2{\left( {{{\tan }^2}{{45}^ \circ } - 3\cos {0^ \circ }} \right)^2}$like $\tan {60^ \circ },\sin {90^ \circ },\tan {45^ \circ }$ and $\cos {0^ \circ }$. Then put these values in the given expression and then simplify it to get some constant value.

Complete step by step solution:
From trigonometric ratios we know
$\tan {60^ \circ } = \sqrt 3 $
$\sin {90^ \circ } = 1$
$\tan {45^ \circ } = 1$
And
$\cos {0^ \circ } = 1$
Now, on putting the values of trigonometric ratios involved into given expression from above values, we get
$
   \Rightarrow \left( {{{\tan }^4}{{60}^ \circ } - {{\sin }^4}{{90}^ \circ }} \right) - 2{\left( {{{\tan }^2}{{45}^ \circ } - 3\cos {0^ \circ }} \right)^2} \\
   \Rightarrow \left\{ {{{\left( {\sqrt 3 } \right)}^4} - {{\left( 1 \right)}^4}} \right\} - 2{\left\{ {{{\left( 1 \right)}^2} - 3\left( 1 \right)} \right\}^2} \\
 $
On simplifying the above expression, we get
\[
   \Rightarrow \left\{ {9 - 1} \right\} - 2{\left\{ {1 - 3} \right\}^2} \\
   \Rightarrow 8 - 2{\left\{ { - 2} \right\}^2} \\
   \Rightarrow 8 - 8 \\
   \Rightarrow 0 \\
\]
Hence, the value of $\left( {{{\tan }^4}{{60}^ \circ } - {{\sin }^4}{{90}^ \circ }} \right) - 2{\left( {{{\tan }^2}{{45}^ \circ } - 3\cos {0^ \circ }} \right)^2}$is zero.

Note:- Whenever you get this type of question the key concept to solve is to learn the different trigonometric ratios like $\sin \theta ,\cos \theta ,\tan \theta ,\cot \theta ,\sec \theta {\text{ and cosec}}\theta $ at different angles like at ${0^ \circ },{30^ \circ },{60^ \circ }{\text{ and 9}}{0^ \circ }$. And one more thing to be noted is that you have to simplify the given expression till some constant value doesn’t come.