Question

1. prove each identity

a) \csc 2x = \frac{\csc x}{2\cos x}
b) \sin x - \tan y \cos x = \frac{\sin(x - y)}{\cos y}

Explanation:

Part (a)

Step 1: Recall reciprocal and double - angle identities

Recall that \(\csc\theta=\frac{1}{\sin\theta}\) and the double - angle identity \(\sin2\theta = 2\sin\theta\cos\theta\). So, \(\csc2x=\frac{1}{\sin2x}\) and \(\csc x=\frac{1}{\sin x}\).
We start with the left - hand side (LHS) of the identity \(\csc2x\).
\(\csc2x=\frac{1}{\sin2x}\)

Step 2: Substitute the double - angle formula for sine

Since \(\sin2x = 2\sin x\cos x\), we substitute this into the expression for \(\csc2x\):
\(\frac{1}{\sin2x}=\frac{1}{2\sin x\cos x}\)

Step 3: Rewrite the expression using the reciprocal identity

We know that \(\frac{1}{\sin x}=\csc x\), so we can rewrite \(\frac{1}{2\sin x\cos x}\) as \(\frac{\csc x}{2\cos x}\), which is the right - hand side (RHS) of the identity.

Part (b)

Step 1: Recall the tangent identity

Recall that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). So, \(\tan y = \frac{\sin y}{\cos y}\).
We start with the left - hand side (LHS) of the identity: \(\sin x-\tan y\cos x\)
Substitute \(\tan y=\frac{\sin y}{\cos y}\) into the expression:
\(\sin x-\frac{\sin y}{\cos y}\cos x\)

Step 2: Simplify the expression

\(\sin x-\frac{\sin y\cos x}{\cos y}=\frac{\sin x\cos y-\sin y\cos x}{\cos y}\)

Step 3: Use the sine subtraction formula

Recall the sine subtraction formula \(\sin(A - B)=\sin A\cos B-\cos A\sin B\). Here, \(A = x\) and \(B = y\), so \(\sin x\cos y-\cos x\sin y=\sin(x - y)\).
So, \(\frac{\sin x\cos y-\sin y\cos x}{\cos y}=\frac{\sin(x - y)}{\cos y}\), which is the right - hand side (RHS) of the identity.

Answer:

a) The identity \(\csc2x=\frac{\csc x}{2\cos x}\) is proven.
b) The identity \(\sin x-\tan y\cos x=\frac{\sin(x - y)}{\cos y}\) is proven.