Verify that each equation is an identity.
csc A sin 2A - sec A = cos 2A sec A

Guida verificata passo dopo passo

Start by writing down the given equation to verify: \(\csc A \sin 2A - \sec A = \cos 2A \sec A\).

Recall the double-angle identity for sine: \(\sin 2A = 2 \sin A \cos A\). Substitute this into the left-hand side (LHS) to get \(\csc A \cdot 2 \sin A \cos A - \sec A\).

Simplify the term \(\csc A \cdot \sin A\) by using the definition \(\csc A = \frac{1}{\sin A}\), so \(\csc A \sin A = 1\). This reduces the LHS to \(2 \cos A - \sec A\).

Express \(\sec A\) as \(\frac{1}{\cos A}\) and rewrite the LHS as \(2 \cos A - \frac{1}{\cos A}\). Find a common denominator to combine the terms into a single fraction.

Now, simplify the right-hand side (RHS) \(\cos 2A \sec A\) by writing \(\sec A = \frac{1}{\cos A}\), so RHS becomes \(\frac{\cos 2A}{\cos A}\). Use the double-angle identity for cosine, \(\cos 2A = 2 \cos^2 A - 1\), and compare the simplified forms of LHS and RHS to verify the identity.

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