Find the Value of \(8\tan x – \sqrt{5}\sec y\)
Question:
If
\[ \sin x = \frac{3}{5}, \quad \tan y = \frac{1}{2} \]
and
\[ \frac{\pi}{2} < x < \pi, \quad \pi < y < \frac{3\pi}{2} \]
find the value of:
\[ 8\tan x – \sqrt{5}\sec y \]
Solution
Given,
\[ \sin x = \frac{3}{5} \]
Since
\[ \frac{\pi}{2} < x < \pi \]
\(x\) lies in Quadrant II, where sine is positive and cosine is negative.
Using:
\[ \sin x = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \]
Take,
\[ \text{Perpendicular} = 3, \quad \text{Hypotenuse} = 5 \]
Using Pythagoras theorem:
\[ \text{Base} = \sqrt{5^2 – 3^2} \]
\[ = \sqrt{25 – 9} = \sqrt{16} = 4 \]
Since \(x\) is in Quadrant II,
\[ \cos x = -\frac{4}{5} \]
Therefore,
\[ \tan x = \frac{\sin x}{\cos x} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \]
Now,
\[ \tan y = \frac{1}{2} \]
Since
\[ \pi < y < \frac{3\pi}{2} \]
\(y\) lies in Quadrant III, where both sine and cosine are negative, but tangent is positive.
Take,
\[ \text{Perpendicular} = 1, \quad \text{Base} = 2 \]
Then,
\[ \text{Hypotenuse} = \sqrt{1^2 + 2^2} = \sqrt{5} \]
In Quadrant III,
\[ \cos y = -\frac{2}{\sqrt{5}} \]
Hence,
\[ \sec y = \frac{1}{\cos y} = -\frac{\sqrt{5}}{2} \]
Now evaluate:
\[ 8\tan x – \sqrt{5}\sec y \]
Substituting the values:
\[ = 8\left(-\frac{3}{4}\right) – \sqrt{5}\left(-\frac{\sqrt{5}}{2}\right) \]
\[ = -6 + \frac{5}{2} \]
\[ = \frac{-12 + 5}{2} \]
\[ = -\frac{7}{2} \]
Final Answer
\[ \boxed{-\frac{7}{2}} \]