The value of

Question:

The value of $\sec ^{2}\left(\tan ^{-1} 2\right)+\operatorname{cosec}^{2}\left(\cot ^{-1} 3\right)$ is ____________________.

Solution:

We know

$\tan ^{-1} x=\sec ^{-1} \sqrt{1+x^{2}}$ and $\cot ^{-1} x=\operatorname{cosec}^{-1} \sqrt{1+x^{2}}$

So,

$\tan ^{-1} 2=\sec ^{-1} \sqrt{1+2^{2}}=\sec ^{-1} \sqrt{5}$

 

$\cot ^{-1} 3=\operatorname{cosec}^{-1} \sqrt{1+3^{2}}=\operatorname{cosec}^{-1} \sqrt{10}$

$\therefore \sec ^{2}\left(\tan ^{-1} 2\right)+\operatorname{cosec}^{2}\left(\cot ^{-1} 3\right)$

$=\sec ^{2}\left(\sec ^{-1} \sqrt{5}\right)+\operatorname{cosec}^{2}\left(\operatorname{cosec}^{-1} \sqrt{10}\right)$

$=\left[\sec \left(\sec ^{-1} \sqrt{5}\right)\right]^{2}+\left[\operatorname{cosec}\left(\operatorname{cosec}^{-1} \sqrt{10}\right)\right]^{2}$

$=(\sqrt{5})^{2}+(\sqrt{10})^{2}$

 

$=15$\

Thus, the value of $\sec ^{2}\left(\tan ^{-1} 2\right)+\operatorname{cosec}^{2}\left(\cot ^{-1} 3\right)$ is 15 .

The value of $\sec ^{2}\left(\tan ^{-1} 2\right)+\operatorname{cosec}^{2}\left(\cot ^{-1} 3\right)$ is ____15____.

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now