1. The problem is to express trigonometric functions \(\sin x\), \(\cos x\), and \(\tan x\) in terms of each other without using inverse functions. 2. Recall the fundamental identities: - \(\tan x = \frac{\sin x}{\cos x}\) - \(\sin^2 x + \cos^2 x = 1\) 3. To express \(\sin x\) in terms of \(\tan x\) and \(\cos x\), use: $$\sin x = \tan x \cdot \cos x$$ 4. To express \(\cos x\) in terms of \(\sin x\) and \(\tan x\), rearrange \(\tan x = \frac{\sin x}{\cos x}\) to get: $$\cos x = \frac{\sin x}{\tan x}$$ 5. To express \(\sin x\) or \(\cos x\) solely in terms of \(\tan x\), use the Pythagorean identity: $$\sin^2 x + \cos^2 x = 1$$ Since \(\tan x = \frac{\sin x}{\cos x}\), let \(\cos x = c\), then \(\sin x = c \tan x\). Substitute into the identity: $$ (c \tan x)^2 + c^2 = 1 \Rightarrow c^2 (\tan^2 x + 1) = 1 \Rightarrow c^2 = \frac{1}{\tan^2 x + 1} $$ Therefore: $$ \cos x = \pm \frac{1}{\sqrt{1 + \tan^2 x}} $$ $$ \sin x = \pm \frac{\tan x}{\sqrt{1 + \tan^2 x}} $$ 6. Summary: - \(\sin x = \tan x \cdot \cos x\) - \(\cos x = \frac{\sin x}{\tan x}\) - \(\cos x = \pm \frac{1}{\sqrt{1 + \tan^2 x}}\) - \(\sin x = \pm \frac{\tan x}{\sqrt{1 + \tan^2 x}}\) This way, you can substitute \(\sin\), \(\cos\), and \(\tan\) with each other without using inverse functions.