We know that, x is in second quadrant and y is in third quadrant.
In second quadrant, cos x and tan x are negative.
In third quadrant, sec y is negative.
By using the formula,
cos x = \( – \sqrt{(1-sin^2 x)}\)
tan x =\(\dfrac{ sin x}{cos x}\)
Now,
cos x = \( – \sqrt{(1-sin^2 x)}\)
= \(\dfrac{ – 4}{5}\)
tan x = \( \dfrac{ sin x}{cos x}\)
= \( \dfrac{3}{5}× \dfrac{-5}{4}\)
= \( \dfrac{-3}{4}\)
We know that \(sec y = – \sqrt{(1+tan^2 y)}\)
= \( \sqrt{-\dfrac{5}{2}}\)
= \(\dfrac{ (-12+5)}{2}\)
=\( \dfrac{-7}{2}\)
\(8 tan x – \sqrt{5} sec y = \dfrac{-7}{2}\)
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