Question

If
$\sin(x) = 1 \div 3 \: \: \tan(y) = 1 \div 2$
and
$(\pi \div 2) < x < \pi < y < 3\pi \div 2$
then
$8 \tan(x) - \sqrt{5 \sec(y) }$
=?​

Answer 1

$\red{ \mathcal{Correct \: Question}}$

If

$\sin(x) = 1 \div 3 \: \: \tan(y) = 1 \div 2$

and

$(\pi \div 2) < x < \pi < y < 3\pi \div 2$

then

$8 \tan(x) - \sqrt{5} \sec(y)$

Answer -->

$\large {\bold {\sin(x) = \frac{1}{3}}} \\ : \to \cos(x) = - \sqrt{1 - { \sin(x) }^{2} } \\ = - \sqrt{ \frac{8}{9 } } \\ : \to \: \tan(x) = \frac{ \frac{ - 1}{3} }{ { \frac{2 \sqrt{2} }{3} } } = \frac{ - 1}{2 \sqrt{2} }$

$\large{\bold{ \tan(y) = \frac{1}{2} }} \\ \\ : \to \: \sec(y) = - \sqrt{1 + \tan(y)^{2} } \\ : \to \: \sec(y) = - \frac{ \sqrt{5} }{2}$

$8 \tan(x) - \sqrt{5} \sec(y) = 4 \times \frac{ - 1}{2 \sqrt{2} } - \sqrt{5} \times \frac{ - \sqrt{5} }{2 } \\ \\ = \bold{ \boxed{\frac{ - 4 \sqrt{2} + 5 }{2} }}$

Hope this Help you