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Question: The value of \[{\sin ^2}\dfrac{{3\pi }}{4} + {\sec ^2}\dfrac{{5\pi }}{3} - {\tan ^2}\dfrac{{2\pi }}{...

The value of sin⁡23π4+sec⁡25π3−tan⁡22π3{\sin ^2}\dfrac{{3\pi }}{4} + {\sec ^2}\dfrac{{5\pi }}{3} - {\tan ^2}\dfrac{{2\pi }}{3} is

Explanation

Solution

Try to express the angles of the trigonometric terms in the form of (nπ±θ)(n\pi \pm \theta ) where n∈Nn \in \mathbb{N} in order to reduce them into some known angles and then put the values of the angles according to their quadrant as the angles are positive or negative in accordance with the 4-quadrant system.

Complete step-by-step answer:
Given, sin⁡23π4+sec⁡25π3−tan⁡22π3{\sin ^2}\dfrac{{3\pi }}{4} + {\sec ^2}\dfrac{{5\pi }}{3} - {\tan ^2}\dfrac{{2\pi }}{3}
Now, we will express all the angles in the form of (nπ±θ)(n\pi \pm \theta )where n∈Nn \in \mathbb{N}
⇒sin⁡2(π−π4)+sec⁡2(2π−π3)−tan⁡2(π−π3)\Rightarrow {\sin ^2}\left( {\pi - \dfrac{\pi }{4}} \right) + {\sec ^2}\left( {2\pi - \dfrac{\pi }{3}} \right) - {\tan ^2}\left( {\pi - \dfrac{\pi }{3}} \right)
Since, sin⁡(π−θ)=sin⁡θ,sec⁡(2π−θ)=sec⁡θ,tan⁡(π−θ)=−tan⁡(θ)\sin (\pi - \theta ) = \sin \theta ,\sec (2\pi - \theta ) = \sec \theta ,\tan \left( {\pi - \theta } \right) = -\tan ( \theta )
⇒sin⁡2(π4)+sec⁡2(π3)−tan⁡2(−π3)\Rightarrow {\sin ^2}\left( {\dfrac{\pi }{4}} \right) + {\sec ^2}\left( {\dfrac{\pi }{3}} \right) - {\tan ^2}\left( { - \dfrac{\pi }{3}} \right)

=(12)2+(2)2−(3)2 =12+4−3 =32  = {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} + {(2)^2} - {\left( {\sqrt 3 } \right)^2} \\\ = \dfrac{1}{2} + 4 - 3 \\\ = \dfrac{3}{2} \\\

Therefore, the value of sin⁡23π4+sec⁡25π3−tan⁡22π3{\sin ^2}\dfrac{{3\pi }}{4} + {\sec ^2}\dfrac{{5\pi }}{3} - {\tan ^2}\dfrac{{2\pi }}{3} is 3/2

Note: These types of questions require the angles of the terms to be reduced into some known angles. The idea of signs of different trigonometric functions in different quadrants is essential.