Question

Find the value of ;
(i) ${\sec }^2\left({\tan }^{-1}2\right)+cose{c}^2\left(co{t}^{-1}3\right)+2$

(ii) $\tan ({\sin }^{-1}\frac{3}{5}+{\cot }^{-1}\frac{3}{2})$

• A. $\left(i\right)=-17,\left(ii\right)=-17/6$
• B. $\left(i\right)=+17,\left(ii\right)=+17/6$
• C. $\left(i\right)=+17,\left(ii\right)-17/6$
• D. $\left(i\right)=-17,\left(ii\right)=+17/6$

Answer 1

The correct option is B $\left(i\right)=+17,\left(ii\right)=+17/6$

i)

${\sec }^2\left({\tan }^{-1}2\right)+cose{c}^2\left({\cot }^{-1}3\right)+2=1+{\tan }^2\left({\tan }^{-1}2\right)+1+{\cot }^2\left({\cot }^{-1}3\right)+2=4+{\left(2\right)}^2+{\left(3\right)}^2=4+4+9=17$
ii)

$\tan ({\sin }^{-1}\frac{3}{5}+{\cot }^{-1}\frac{3}{2})=\tan ({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{3}{2})$
$=\tan \left({\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times \frac{2}{3}}\right)\right)=\tan \left({\tan }^{-1}\left(\frac{17}{6}\right)\right)=\frac{17}{6}$