Question

Show that: $2\left( {co{s^4}{{60}^ \circ } + si{n^4}{{30}^ \circ }} \right) - \left( {ta{n^2}{{60}^ \circ } + co{t^2}{{45}^ \circ }} \right) + 3\;se{c^2}{30^ \circ } = 41$

Answer 1

Here, we are given a trigonometric equation. Next, we will make the table for trigonometric ratios of general angles i.e. ${0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }$. From that, we will find the required values which are there in the given trigonometric equation. On substituting these values and solving them, we will get the final output.

Complete step by step answer:
Given that, $2\left( {co{s^4}{{60}^ \circ } + si{n^4}{{30}^ \circ }} \right) - \left( {ta{n^2}{{60}^ \circ } + co{t^2}{{45}^ \circ }} \right) + 3\;se{c^2}{30^ \circ } = 41$
We need to find the values of all the trigonometric ratios that are there in the equation
i.e. $sin{30^ \circ },cos{60^ \circ },tan{60^ \circ },cot{45^ \circ },sec{30^ \circ }$.
Also, we know that, $\pi = {180^ \circ }$.

First, we will make a table for all general angles of the trigonometric ratios.Here,
$\dfrac{\pi }{6} = {30^ \circ }$, $\dfrac{\pi }{4} = {45^ \circ }$, $\dfrac{\pi }{3} = {60^ \circ }$ and $\dfrac{\pi }{2} = {90^ \circ }$
We will solve the LHS part of this equation and then compare it with the RHS part as shown below:We will substitute all the values using the above table as shown.
$LHS = 2\left( {co{s^4}{{60}^ \circ } + si{n^4}{{30}^ \circ }} \right) - \left( {ta{n^2}{{60}^ \circ } + co{t^2}{{45}^ \circ }} \right) + 3\;se{c^2}{30^ \circ }$

Substituting the value of $cos{60^ \circ } = \dfrac{1}{2}$ and $sin{30^ \circ } = \dfrac{1}{2}$ , we will get,
$LHS = 2\left( {{{\left( {\dfrac{1}{2}} \right)}^4} + {{\left( {\dfrac{1}{2}} \right)}^4}} \right) - \left( {ta{n^2}{{60}^ \circ } + co{t^2}{{45}^ \circ }} \right) + 3\;se{c^2}{30^ \circ }$
Substituting the value of $tan{60^ \circ } = \sqrt 3$ and $cot{45^ \circ } = 1$ , we will get,
$LHS = 2\left( {{{\left( {\dfrac{1}{2}} \right)}^4} + {{\left( {\dfrac{1}{2}} \right)}^4}} \right) - \left( {{{\left( {\sqrt 3 } \right)}^2} + {1^2}} \right) + 3\;se{c^2}{30^ \circ }$
Substituting the value of $sec{30^ \circ } = \dfrac{2}{{\sqrt 3 }}$ , we will get,
$LHS = 2\left( {{{\left( {\dfrac{1}{2}} \right)}^4} + {{\left( {\dfrac{1}{2}} \right)}^4}} \right) - \left( {{{\left( {\sqrt 3 } \right)}^2} + {1^2}} \right) + 3\;{\left( {\dfrac{2}{{\sqrt 3 }}} \right)^2}$

On evaluating this and removing the brackets, we will get,
$LHS = 2\left( {\dfrac{1}{{16}} + \dfrac{1}{{16}}} \right) - \left( {3 + 1} \right) + 3\;\left( {\dfrac{4}{3}} \right)$
Taking LCM of the first term, we will get,
$LHS = 2\left( {\dfrac{{1 + 1}}{{16}}} \right) - \left( 4 \right) + 4$
$\Rightarrow LHS = 2\left( {\dfrac{2}{{16}}} \right) - 4 + 4$
$\Rightarrow LHS = 2\left( {\dfrac{1}{8}} \right) + 0$
Again simplify more, we will get,
$LHS = \dfrac{1}{4} =RHS$

Hence, the given equation is proved.

Note: In these types of questions, students just need to remember the values of $\sin \theta$ and $\cos \theta$ for all the general angles i.e. ${0^ \circ },{\text{ }}{30^ \circ },{\text{ }}{45^ \circ },{\text{ }}{60^ \circ },{\text{ }}{90^ \circ }$. Then from that, they will get the value of $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$. Like for example, to get the value of $\tan {30^ \circ }$, we need to substitute the values of $\sin {30^ \circ }$ and $\cos {30^ \circ }$ and so on. All the other trigonometric ratios are inverse of the sin, cos and tan. $\cos ec\theta = \dfrac{1}{{\sin \theta }}$ , $sec\theta = \dfrac{1}{{\cos \theta }}$ and $\cot \theta = \dfrac{1}{{\tan \theta }}$. Trigonometry is divided into two sub-branches i.e. plane trigonometry and spherical geometry.