Question

The value of ${\sec }^2\left({\tan }^{-1}2\right)+{\text{cosec}}^2\left({\cot }^{-1}3\right)=$ ____

• A. $6$
• B. $15$
• C. $13$
• D. $25$

Answer 1

The correct option is B $15$

Let ${\tan }^{-1}2=x$ (1)

$2=\tan x$

Let

$perpendicular=p=2k$

$base=b=k$

$hypotenuse=h=\sqrt{(2k{)}^2+k^2}=\sqrt{(5k{)}^2}=\sqrt{5}k$

$\sec x=\frac{hypotenuse}{base}=\frac{\sqrt{5}k}{k}=\sqrt{5}$

$x=se{c}^{-1}\sqrt{5}$ (2)

Let ${\cot }^{-1}3=y$ (3)

$3=\cot y$

Let

$base=3=k^{{}^{\prime }}$

$perpendicular=p=k^{{}^{\prime }}$

$hypotenuse=h=\sqrt{(k^{{}^{\prime }}{)}^2+(3k^{{}^{\prime }}{)}^2}=\sqrt{(10k^{{}^{\prime }}{)}^2}=\sqrt{10}{k}^{{}^{\prime }}$

$\text{cosec}y=\frac{hypotenuse}{perpendicular}=\frac{\sqrt{10}{k}^{{}^{\prime }}}{k^{{}^{\prime }}}=\sqrt{10}$

$y={\text{cosec}}^{-1}\sqrt{10}$ (4)

From (1),(2),(3) and (4)

${\sec }^2\left({\tan }^{-1}2\right)+{\text{cosec}}^2\left({\cot }^{-1}3\right)={\sec }^{2}x+{\text{cosec}}^{2}y={\sec }^{2}se{c}^{-1}\sqrt{5}+{\text{cosec}}^2{\text{cosec}}^{-1}\sqrt{10}$

$=(\sec {\sec }^{-1}\sqrt{5}{)}^2+(\text{cosec}{\text{cosec}}^{-1}\sqrt{10}{)}^2=(\sqrt{5}{)}^2+(\sqrt{10}{)}^2=5+10=15$