Question

Find the value of $4{\sin ^2}60^\circ + 3{\tan ^2}30^\circ - 8\sin 45^\circ .\cos 45^\circ$.

Answer 1

Hint : The given problem revolves around the concepts of trigonometric ratios. Here, we are given a trigonometric expression $4{\sin ^2}60^\circ + 3{\tan ^2}30^\circ - 8\sin 45^\circ .\cos 45^\circ$ and we have to find the value of it. As we can see we are given various trigonometric ratios for any angle, we just need to substitute the values of trigonometric ratios of a particular angle and add or subtract them according to the given trigonometric expression.

Complete step by step solution:
We have,
$4{\sin ^2}60^\circ + 3{\tan ^2}30^\circ - 8\sin 45^\circ .\cos 45^\circ$
We can solve the given trigonometric expression by calculating the values of various trigonometric ratios and then substituting them in the given trigonometric expression.
We can see that $\sin e$ and $\tan gent$ functions are given in the form of squares.
${\sin ^2}60^\circ$ is nothing but square of $\sin 60^\circ$ i.e., ${\left( {\sin 60^\circ } \right)^2}$.
$\Rightarrow {\sin ^2}60^\circ = {\left( {\sin 60^\circ } \right)^2}$
As we know the value of $\sin 60^\circ$is $\left( {\dfrac{{\sqrt 3 }}{2}} \right)$ .
$\Rightarrow {\left( {\sin 60^\circ } \right)^2} = {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2}$
$\Rightarrow {\left( {\sin 60^\circ } \right)^2} = \left( {\dfrac{3}{4}} \right)$
Similarly we can write ${\tan ^2}30^\circ$ as${\left( {\tan 30^\circ } \right)^2}$.
$\Rightarrow {\tan ^2}30^\circ = {\left( {\tan 30^\circ } \right)^2}$
As we know the value of $\tan 30^\circ$ is$\left( {\dfrac{1}{{\sqrt 3 }}} \right)$.
$\Rightarrow {\left( {\tan 30^\circ } \right)^2} = {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2}$
$\Rightarrow {\left( {\tan 30^\circ } \right)^2} = \left( {\dfrac{1}{3}} \right)$
As we know value of $\sin 45^\circ$is $\left( {\dfrac{1}{{\sqrt 2 }}} \right)$ and value of $\cos 45^\circ$ is $\left( {\dfrac{1}{{\sqrt 2 }}} \right)$.
Now we know trigonometric expression $4{\sin ^2}60^\circ + 3{\tan ^2}30^\circ - 8\sin 45^\circ .\cos 45^\circ$ can also be written as $4{\left( {\sin 60^\circ } \right)^2} + 3{\left( {\tan 30^\circ } \right)^2} - 8\sin 45^\circ .\cos 45^\circ$. Now we will substitute the values of trigonometric ratios of various angles we have find above in $4{\left( {\sin 60^\circ } \right)^2} + 3{\left( {\tan 30^\circ } \right)^2} - 8\sin 45^\circ .\cos 45^\circ$.
$4{\left( {\sin 60^\circ } \right)^2} + 3{\left( {\tan 30^\circ } \right)^2} - 8\sin 45^\circ .\cos 45^\circ$
Substitute the values of trigonometric ratios.
$\Rightarrow 4 \times \left( {\dfrac{3}{4}} \right) + 3 \times \left( {\dfrac{1}{3}} \right) - 8 \times \left( {\dfrac{1}{{\sqrt 2 }}} \right) \times \left( {\dfrac{1}{{\sqrt 2 }}} \right)$
Opening brackets, we get
$\Rightarrow \dfrac{{4 \times 3}}{4} + \dfrac{{3 \times 1}}{3} - \dfrac{{8 \times 1}}{2}$
$\Rightarrow \dfrac{3}{1} + \dfrac{1}{1} - \dfrac{8}{2}$
After simplifying, we get
$\Rightarrow 3 + 1 - 4$
$\Rightarrow 4 - 4$
$= 0$
Therefore, the value of $4{\sin ^2}60^\circ + 3{\tan ^2}30^\circ - 8\sin 45^\circ .\cos 45^\circ$is$0$.
So, the correct answer is “0”.

Note : Take care of the values of trigonometric ratios you are substituting. For solving these types of questions you need to remember all the values of trigonometric ratios. Remember to solve the problem using BODMAS (brackets, of, division, multiplication, addition, and subtraction). Check your calculations.