Question

Find the value of expression.
$\displaystyle \sf 3 \ tan^2 \ \dfrac{\pi}{ 6} \ + \ \dfrac{4}{3} \ cos^2 \ \dfrac{\pi}{6} \ - \ \dfrac{1}{2} \ cot^3 \ \dfrac{\pi}{4} - \dfrac{2}{3} \ sin^2 \ \dfrac{\pi}{3} \ +\ \dfrac{1}{8} \ sec^4 \ \dfrac{\pi}{3}$

Answer 1

⭐《ANSWER》

$\huge\mathfrak\pink {ANSWER}$

↪Actually welcome to the concept of the TRIGONOMETRIC EQUATIONS

↪After solving we get Answer as 3

Answer 2

Step-by-step explanation:

$Given: 3 tan^2 \frac{\pi }{6} +\frac{4}{3} cos^2\frac{\pi }{6} - \frac{1}{2} cot^3 \frac{\pi }{4} -\frac{2}{3} sin^2\frac{\pi }{3} +\frac{1}{8} sec^4 \frac{\pi }{3}$

$\Longrightarrow 3(\frac{1}{\sqrt{3}})^2+\frac{4}{3}(\frac{\sqrt{3}}{2})^2 - \frac{1}{2}(1)^3-\frac{2}{3}(\frac{\sqrt{3}}{2})^2 +\frac{1}{8}(\frac{{1}}{\frac{1}{2}})^4$

$\Longrightarrow 1 +\frac{4}{3}(\frac{3}{4})-\frac{1}{2}(1)-\frac{2}{3}(\frac{3}{4}) +\frac{1}{8}(16)$

$\Longrightarrow 1 + 1 - \frac{1}{2} - \frac{1}{2} + 2$

$\Longrightarrow 1 + 1 - 1 + 2$

$\Longrightarrow \textbf 3$

Hope it helps!