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.