Question

The value of ${\sin ^2}\dfrac{{3\pi }}{4} + {\sec ^2}\dfrac{{5\pi }}{3} - {\tan ^2}\dfrac{{2\pi }}{3}$ is

Answer 1

Hint: We have 3 trigonometric functions. We can convert the angles to sum or difference of$\pi $. Then we can find their values using the equations $\sin \left( {\pi - x} \right) = \sin x$, $\sec \left( {2\pi - x} \right) = \sec x$ and $\tan \left( {\pi - x} \right) = - \tan x$. Then we can square these values and substitute in the given expression to get the required answer.

Complete step by step answer:

We have the expression ${\sin ^2}\dfrac{{3\pi }}{4} + {\sec ^2}\dfrac{{5\pi }}{3} - {\tan ^2}\dfrac{{2\pi }}{3}$. We can find the value of each term separately and add them together to get the required value of the expression.
Let us take the 1st term, ${\sin ^2}\dfrac{{3\pi }}{4}$.
The angle $\dfrac{{3\pi }}{4}$ can be written as $\pi - \dfrac{\pi }{4}$.
$ \Rightarrow \sin \dfrac{{3\pi }}{4} = \sin \left( {\pi - \dfrac{\pi }{4}} \right)$
We know that, $\sin \left( {\pi - x} \right) = \sin x$. Applying this, we get,
$ \Rightarrow \sin \dfrac{{3\pi }}{4} = \sin \left( {\dfrac{\pi }{4}} \right)$
We know that $\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$.
$ \Rightarrow \sin \dfrac{{3\pi }}{4} = \dfrac{1}{{\sqrt 2 }}$
Taking the square, we get,
$ \Rightarrow {\sin ^2}\dfrac{{3\pi }}{4} = \dfrac{1}{2}$ … (1)
Now we can take the 2nd term, ${\sec ^2}\dfrac{{5\pi }}{3}$
The angle $\dfrac{{5\pi }}{3}$can be written as $2\pi - \dfrac{\pi }{3}$.
$ \Rightarrow \sec \dfrac{{5\pi }}{3} = \sec \left( {2\pi - \dfrac{\pi }{3}} \right)$
We know that, $\sec \left( {2\pi - x} \right) = \sec x$
$ \Rightarrow \sec \dfrac{{5\pi }}{3} = \sec \left( {\dfrac{\pi }{3}} \right)$
We know that $\sec \left( {\dfrac{\pi }{3}} \right) = 2$
$ \Rightarrow \sec \dfrac{{5\pi }}{3} = 2$
Taking the square, we get,
$ \Rightarrow {\sec ^2}\dfrac{{5\pi }}{3} = 4$ … (2)
Now we can take the 3rd term, ${\tan ^2}\dfrac{{2\pi }}{3}$
The angle $\dfrac{{2\pi }}{3}$can be written as $\pi - \dfrac{\pi }{3}$.
$ \Rightarrow \tan \dfrac{{2\pi }}{3} = \tan \left( {\pi - \dfrac{\pi }{3}} \right)$
We know that,$\tan \left( {\pi - x} \right) = - \tan x$
$ \Rightarrow \tan \dfrac{{2\pi }}{3} = - \tan \left( {\dfrac{\pi }{3}} \right)$
We know that $\tan \left( {\dfrac{\pi }{3}} \right) = \sqrt 3 $
$ \Rightarrow \tan \dfrac{{2\pi }}{3} = - \sqrt 3 $
Taking the square, we get,
$ \Rightarrow {\tan ^2}\dfrac{{2\pi }}{3} = 3$ … (3)
On substituting equations (1), (2) and (3) in the given expression, we get,
${\sin ^2}\dfrac{{3\pi }}{4} + {\sec ^2}\dfrac{{5\pi }}{3} - {\tan ^2}\dfrac{{2\pi }}{3}$
$ = \dfrac{1}{2} + 4 - 3$
$ = \dfrac{1}{2} + 1 = \dfrac{3}{2}$
Therefore the value of the given expression is $\dfrac{3}{2}$

Note: We must know the values of trigonometric functions at common angles. Adding $\pi $or multiples of $\pi $with the angle retains the ratio and adding $\dfrac{\pi }{2}$or odd multiples of $\dfrac{\pi }{2}$will change the ratio. While converting the angles we must take care of the sign of the ratio in its respective quadrant. In the 1^st quadrant, all the trigonometric ratios are positive. In the 2^nd quadrant, only sine and sec are positive. In the third quadrant, only tan and cot are positive and in the fourth quadrant, only cos and sec are positive. The following figure gives us an idea about the signs of different trigonometric functions. The angle measured in the counterclockwise direction is taken as positive and the angle measured in the clockwise direction is taken as negative.