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  4. The value of tan2(sec-12) + cot2(cosec-13) is

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Q. The value of $tan^2(sec^{-1}2) + cot^2(cosec^{-1}3)$ is

Inverse Trigonometric Functions

Solution:

$tan^2 (sec^{-1} 2) + cot^2(cosec^{-1} 3)$
$ = tan^2 t_1 + cot^2 \,t_2, $ where $t_1 = sec^{-1} 2, t_2 = cosec^{-1} 3$
' $=sec^2 t - 1 + cosec^2 t_2 - 1$
$= sec^2 t_1 + cosec^2 t_2 - 2 = 4 + 9 -2 = 11$
Alternative Solution :
$f(x) = \begin{cases} Let & t_1 = sec^{-1} 2 \\ \therefore &sec\,t_1 = 2 \\ & 1+ tan^2 t_1 = 4\\ & tan^2 t_1 = 3 \end{cases} $ Similarly $ \begin{cases} t_2 = cosec^{-1} 3 \\ cosec\, t_2 = 3 \\ cosec^2\,t_2 =9 \\ cot^2 t_2 = 8 \end{cases} $
$ \therefore tan^2 t_1 + cot^2 t_2 = 3 + 8 = 11$
Short Cut Method :
$tan^2 (sec^{-1} 2) + cot^2(cosec^{-1} 3)$
$ = -1+ [ sec^2(sec^{-1} 2)] + cosec^2(cosec^{-1} 3 ) - 1$
$ = 4 - 1 + 9 - 1 = 11$