If sin θ = 3/5, tan φ = 1/2 and π/2 < θ < π < φ < 3π/2, then find the value of 8 tan θ – √5 sec φ.

Question

If sin θ = \(\frac{3}{5}\), tan φ = \(\frac{1}{2}\) and \(\frac{\pi}{2}\)< θ < π < φ < \(\frac{3\pi}{2}\), then find
the value of 8 tan θ – √5 sec φ.

Answer 1

Given that sin θ = \(\frac{3}{5}\) = \(\frac{opposite\,side}{Hypotenuse}\)

∵ AB = \(\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}\) = 4

Here θ lies in second quadrant [∵ \(\frac{\pi}{2}\) < θ < π] 

∵ tan θ is negative. 

tan θ = \(-\frac{3}{4}\)

Also given that tan Φ = \(\frac{1}{2}\) = \(\frac{opposite\,side}{adjacent\,side}\)

∴ PR = \(\sqrt{PQ^2+QP^2}=\sqrt{4+1}=\sqrt{5}\)

Here Φ lies in third quadrant (∵ π < Φ < \(\frac{3\pi}{2}\)) 

∴ sec Φ is negative.

sec Φ = \(\frac{1}{cosΦ}=-\frac{1}{\frac{2}{\sqrt{5}}}=-\frac{\sqrt{5}}{2}\)

Now 8 tan θ – √5 sec Φ = 8(-\(\frac{3}{4}\)) - \(\sqrt{5}\)(\(\frac{-\sqrt{5}}{2}\))

= 2 x (-3) + \(\frac{5}{2}\)

= -6 + \(\frac{5}{2}\)

= \(\frac{-12+5}{2}\)

= \(\frac{-7}{2}\)