Question

If sin\[x = \frac{3}{5}, \tan y = \frac{1}{2}\text{ and }\frac{\pi}{2} < x < \pi < y < \frac{3\pi}{2},\]  find the value of 8 tan \[x - \sqrt{5} \sec y\]

Answer 1

We have:
\[\sin x = \frac{3}{5}, \tan y = \frac{1}{2}\text{ and }\frac{\pi}{2} < x < \pi < y < \frac{3\pi}{2}, \]
Thus, x is in the second quadrant and y is in the third quadrant. 
\[\text{ In the second quadrant, }\cos x \text{ and }\tan x \text{are negative.} \]
In the third quadrant, secy is negative .
\[\therefore \cos x = - \sqrt{1 - \sin^2 x} = - \sqrt{1 - \left( \frac{3}{5} \right)^2} = \frac{- 4}{5}\]
\[\tan x = \frac{\frac{3}{5}}{\frac{- 4}{5}} = \frac{- 3}{4}\]
\[\text{ And, }\sec y = - \sqrt{1 + \tan^2 y} = - \sqrt{1 + \left( \frac{1}{2} \right)^2} = \frac{- \sqrt{5}}{2}\]
\[ \therefore 8 \tan x - \sqrt{5} \sec y = 8 \times \frac{- 3}{4} - \sqrt{5} \times \frac{- \sqrt{5}}{2} = - 6 + \frac{5}{2} = - \frac{7}{2}\]