Question

Determinati x apartine cadranului (0,π/2) pentru care (1+sinx)/sinx=(1+cosx)/cosx

Answer 1

(1+sinx)/sinx=(1+cosx)/cosx
cosx+cosxsinx=sinx+cosxsinx
cosx-sinx=0
1-sinx/cosx=0
1-tagx=0
tagx=1
x=$\pi /4+ \pi n$

Answer 2

$x\in \Big(0,\dfrac{\pi}{2}\Big) \\ \\ \dfrac{1+\sin x}{\sin x} = \dfrac{1+\cos x}{\cos x} \Rightarrow \dfrac{1}{\sin x}+\dfrac{\sin x}{\sin x} = \dfrac{1}{\cos x}+\dfrac{\cos x}{\cos x} \Rightarrow \\ \\ \Rightarrow \dfrac{1}{\sin x}+1 = \dfrac{1}{\cos x} +1 \Rightarrow \dfrac{1}{\sin x} = \dfrac{1}{\cos x} \Rightarrow \cos x = \sin x \Big|:\cos x\Rightarrow \\ \\ (Putem imparti prin cosx deoarece cosx nu ia valoarea 0 cand x \\ apartine \Big(0,\dfrac{\pi}{2}\Big) )$

$\Rightarrow 1 = \dfrac{\sin x}{\cos x} \Rightarrow \dfrac{\sin x}{\cos x} =1 \Rightarrow \text{tg} x = 1 \Rightarrow x = \dfrac{\pi}{4} +k\pi \\ \\ \bullet k = -1 \Rightarrow x = \dfrac{\pi}{4}-\pi \notin \Big(0,\dfrac{\pi}{2}\Big) \\ \\ \bullet k = 0 \Rightarrow x = \dfrac{\pi}{4} \in \Big(0,\dfrac{\pi}{2}\Big) \\ \\ \bullet k = 1 \Rightarrow x = \dfrac{\pi}{4}+\pi \Rightarrow x = \dfrac{\pi+4\pi}{4} \Rightarrow x = \dfrac{5\pi}{4} \in \Big(0,\dfrac{\pi}{2}\Big)$
$\bullet k = 2 \Rightarrow x = \dfrac{\pi}{4}+2\pi \Rightarrow x = \dfrac{\pi+8\pi}{4} \Rightarrow x = \dfrac{9\pi}{4}\notin \Big(0,\dfrac{\pi}{2}\Big)\\ \\ \Rightarrow S = \Big\{\dfrac{\pi}{4},\dfrac{5\pi}{4}\Big\}$