Question

Find the value of
$ta{n}^2\left(se{c}^{-1}\left(3\right)\right)$+$co{t}^2\left(cose{c}^{-1}\left(4\right)\right)$

• A. $5$
• B. $25$
• C. $7$
• D. $23$

Answer 1

The correct option is D $23$

We know that
${\sec }^2\theta -{\tan }^2\theta =1\mathrm{and}{\mathrm{cosec}}^2\theta -{\cot }^2\theta =1.\therefore {\tan }^2\left({\sec }^{-1}3\right)+{\cot }^2\left({\mathrm{cosec}}^{-1}4\right)={\sec }^2\left({\sec }^{-1}3\right)-1+{\mathrm{cosec}}^2\left({\mathrm{cosec}}^{-1}4\right)-1={\left(\sec \left({\sec }^{-1}3\right)\right)}^2+{\left(\mathrm{cosec}\left({\mathrm{cosec}}^{-1}4\right)\right)}^2-2$

$\mathrm{Since},\sec \left({\sec }^{-1}\theta \right)=\theta \mathrm{and}\mathrm{cosec}\left({\mathrm{cosec}}^{-1}\theta \right)=\theta \mathrm{for}\mathrm{all}\theta \in (-\infty ,-1]\cup [1,\infty )\mathrm{Therefore},{\tan }^2\left({\sec }^{-1}3\right)+{\cot }^2\left({\mathrm{cosec}}^{-1}4\right)={\left(3\right)}^2+{\left(4\right)}^2-2=23.$