91Ó°ÊÓ

Recall that one of the main trigonometric Pythagorean identities is: \[ 1 = \sin^2 x + \cos^2 x \] We will manipulate this identity to rewrite the given expression. Subtract \( \cos^2 x \) from both sides of the Pythagorean identity: \[ 1 - \cos^2 x = \sin^2 x \] Now, let's go to the given expression: \[ \frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x} \] Let's focus on the left-hand side (LHS): \[ \frac{\sin x}{1-\cos x} \] We will use the identity that we just derived to rewrite the denominator as follows: \[ \frac{\sin x}{\sin^2 x} \]

Continuing from the last step, we have the following expression: \[ \frac{\sin x}{\sin^2 x} \] Next, notice that we can simplify this expression by canceling out a common \(\sin x\) term from both the numerator and the denominator. Cancel out the common term \(\sin x\): \[ \frac{1}{\sin x} \] This is our simplified expression.

Now we need to check whether this simplified expression is equal to the right-hand side (RHS) of the given exercise: \[ \frac{1}{\sin x} \stackrel{?}{=} \frac{1+\cos x}{\sin x} \] Notice that we cannot further simplify the RHS of the given expression, and our simplified expression does not immediately match the RHS. But if we recall the manipulation from Step 1, we can easily see that if we multiply the numerator and denominator of the RHS by \((1 - \cos x)\), the resulting expression will be equal to our simplified expression: \[ \frac{(1+\cos x)(1-\cos x)}{\sin x(1-\cos x)} \] Which simplifies to: \[ \frac{1 - \cos^2 x}{\sin x} \] Using the trigonometric identity from Step 1 again, we can substitute \(\sin^2 x\) for \(1 - \cos^2 x\): \[ \frac{\sin^2 x}{\sin x} \] Now, this clearly matches our simplified expression from Step 2, and thus: \[ \frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x} \] We have successfully proven the required trigonometric identity.