Question

Show that f(x) = tan⁻¹ (sin x + cos x) is a decreasing function on the interval (π/4, π/2).

Answer 1

$$\begin{gathered} f\left(x\right)={\tan }^{-1}\left(\sin x+\cos x\right) \\ f\text{'}\left(x\right)=\frac{1}{1+{\left(\sin x+\cos x\right)}^2}\left(\cos x-\sin x\right) \\ =\frac{1}{1+1+2\sin x\cos x}\left(\cos x-\sin x\right) \\ =\frac{\left(\cos x-\sin x\right)}{2+\sin 2x} \\ \text{Here,} \\ \frac{\mathrm{\pi }}{4}<x<\frac{\mathrm{\pi }}{2} \\ \Rightarrow \frac{\mathrm{\pi }}{2}<2x<\mathrm{\pi } \\ \Rightarrow \sin 2x>0 \\ \Rightarrow 2+\sin 2x>0...\left(1\right) \\ \mathrm{Also}, \\ \frac{\mathrm{\pi }}{4}<x<\frac{\mathrm{\pi }}{2} \\ \cos x<\sin x \\ \Rightarrow \cos x-\sin \mathit{}x<0...\left(2\right) \\ f\text{'}\left(x\right)=\frac{\left(\cos x-\sin x\right)}{2+\sin 2\mathrm{x}}<0,\forall x\in \left(\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right)\left[\mathrm{From}\mathrm{eqs}.(1)\; and\; (2)\right] \\ \text{So,}f\left(x\right)\text{is decreasing on}\left(\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{2}\right)\text{.} \end{gathered}$$