Question

The value of $ta{n}^2(se{c}^{-1}2)+co{t}^2(cose{c}^{-1}3)$ is

• A. $15$
• B. $13$
• C. $11$
• D. None of these

Answer 1

Answer: (C)

$ta{n}^2(se{c}^{-1}2)+co{t}^2(cose{c}^{-1}3)$
$=ta{n}^2{t}_1+co{t}^2{t}_2,$ where $t_1=se{c}^{-1}2,t_2=cose{c}^{-1}3$
' $=se{c}^{2}t-1+cose{c}^2{t}_2-1$
$=se{c}^2{t}_1+cose{c}^2{t}_2-2=4+9-2=11$
Alternative Solution :
$f(x)=\{\begin{array}{ll}Let& t_1=se{c}^{-1}2\\ \therefore & sec{t}_1=2\\ & 1+ta{n}^2{t}_1=4\\ & ta{n}^2{t}_1=3\end{array}$ Similarly $\{\begin{array}{l}{t}_2=cose{c}^{-1}3\\ cosec{t}_2=3\\ cose{c}^2{t}_2=9\\ co{t}^2{t}_2=8\end{array}$
$\therefore ta{n}^2{t}_1+co{t}^2{t}_2=3+8=11$
Short Cut Method :
$ta{n}^2(se{c}^{-1}2)+co{t}^2(cose{c}^{-1}3)$
$=-1+[se{c}^2(se{c}^{-1}2)]+cose{c}^2(cose{c}^{-1}3)-1$
$=4-1+9-1=11$