If \(\tan x=\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta} \), \(\frac{\pi}{4}<\theta<\frac{\pi}{2} \), then what is √2 sin x equal to?

Given:

tan x = \(\frac{\sin θ+\cos θ}{\sin θ-\cos θ}\), \(\frac{\pi}{4}<θ<\frac{\pi}{2} \)

Formula Used:

1. tan θ = \(\frac{P}{B}\)

2. sin θ = \(\frac{P}{H}\) 

3. sin²θ + cos²θ = 1

Calculation:

By Pythagoras Theorem

⇒ H = \(\sqrt{(sinθ+cosθ)^2+(sinθ-cosθ)^2} \)

⇒ H = \(\sqrt{sin^2θ+cos^2θ+2sinθ cosθ+sin^2θ+cos^2θ-2sinθ cosθ} \)

⇒ H = \(\sqrt{2(sin^2θ+cos^2θ)} \) = \(\sqrt{2} \)

According to the figure

⇒ \(\sqrt{2} \) sin x = \(\sqrt{2} \) × \(\frac{sinθ+\cosθ}{\sqrt{2}}\) = sinθ + cosθ 

∴ The value of \(\sqrt{2} \) sin x is equal to sinθ + cosθ.