Answer:

the answer is

[tex] \frac{3}{2} [/tex]

Step-by-step explanation:

according to the question,

[tex] \sin^{2} \frac{3\pi}{4} + \sec ^{2} \frac{5\pi}{3} + \tan^{2} \frac{2\pi}{3} [/tex]

now, it can be written as

[tex] \sin^{2} \frac{3\pi}{4} =sin^{2}(\pi - \frac{\pi}{4} )\\ \sec ^{2} \frac{5\pi}{3} =sec ^{2}(2\pi - \frac{\pi}{3} ) \\ \tan^{2} \frac{2\pi}{3} = tan^{2}(\pi - \frac{\pi}{ 3} )[/tex]

Now as we know that in 1st quadrant ( 0° - 90°) all trigonometric functions are positive

in 2nd quadrant (90° - 180°) only sin and cosec is positive rest all trigonometric functions are negative.

in 3rd quadrant (180° - 270°) only tan and cot is positive rest all trigonometric functions are negative.

in 4th quadrant (270° - 360°) only cos and sec is positive rest all trigonometric functions are negative.

therefore,

[tex] \sin^{2} \frac{3\pi}{4} =sin^{2}(\pi - \frac{\pi}{4} ) \\ =sin^{2}( \frac{\pi}{4} ) \\ = { (\frac{1}{ \sqrt{2} } })^{2} \\ = \frac{1}{ 2} [/tex]

since,

[tex] \sin(\pi - x) = \sin(x) \\ \sec(2\pi - x) = \sec(x) \\ \tan(\pi - x) = - \tan(x) [/tex]

now,

[tex]\sec ^{2} \frac{5\pi}{3} =sec ^{2}(2\pi - \frac{\pi}{3} ) \\ = \sec^{2} ( \frac{\pi}{3} ) \\ = {2}^{2} = 4[/tex]

[tex]\tan^{2} \frac{2\pi}{3} = tan^{2}(\pi - \frac{\pi}{ 3} ) \\ = - \tan^{2} ( \frac{\pi}{3} ) \\ = - ( {\sqrt{3} })^{2} \\ = - 3[/tex]

therefore,

[tex] \sin^{2} \frac{3\pi}{4} + \sec ^{2} \frac{5\pi}{3} + \tan^{2} \frac{2\pi}{3} = \frac{1}{2} + 4 - 3 \\ = \frac{1 + 8 - 6}{2} \\ = \frac{3}{2} [/tex]

hope it helps