Question

${\displaystyle {\sin }^2\left({\sin }^{-1} \frac{1}{2}\right)+{\tan }^2\left({\sec }^{-1} 2\right)+{\cot }^2(\cos e{c}^{-1} 4)}$ = ______.

Options

• ${\displaystyle \frac{73}{4}}$

• ${\displaystyle \frac{37}{2}}$

• ${\displaystyle \frac{89}{4}}$

• 19

Answer 1

${\displaystyle {\sin }^2\left({\sin }^{-1} \frac{1}{2}\right)+{\tan }^2\left({\sec }^{-1} 2\right)+{\cot }^2(\cos e{c}^{-1} 4)=\underline{\frac{73}{4}}}$.

Explanation:

Let cosec^-1 4 = θ ⇒ cosec θ = 4

cot² θ = cosec² θ - 1 = (4)² - 1 = 15

${\displaystyle {\sin }^2\left({\sin }^{-1} \frac{1}{2}\right)+{\tan }^2\left({\sec }^{-1} 2\right)+{\cot }^2(\cos e{c}^{-1} 4)}$

${\displaystyle ={\sin }^2{30}^{\circ }+{\tan }^2{60}^{\circ }+{\cot }^2\theta }$

${\displaystyle ={\left(\frac{1}{2}\right)}^2+{\left(\sqrt{3}\right)}^2+15}$

${\displaystyle =\frac{73}{4}}$