Proving a Reciprocal Trig Identity

To prove an identity involving $\sec$ and $\csc$, rewrite them as $\frac{1}{\cos}$ and $\frac{1}{\sin}$, put everything over a common denominator, and use $\sin^2 x + \cos^2 x = 1$. Work one side only until it becomes the other.

Prove that $\sec^2 x + \csc^2 x = \sec^2 x\,\csc^2 x$. [4]

The working

Step 1, start with the left-hand side and write in terms of $\sin$ and $\cos$: $\text{LHS} = \sec^2 x + \csc^2 x = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \quad \text{(M1)}$

Step 2, combine over the common denominator $\sin^2 x\cos^2 x$: $= \frac{\sin^2 x + \cos^2 x}{\sin^2 x\,\cos^2 x} \quad \text{(M1)}$

Step 3, use $\sin^2 x + \cos^2 x = 1$ in the numerator: $= \frac{1}{\sin^2 x\,\cos^2 x} \quad \text{(A1)}$

Step 4, split back into the reciprocal squares: $= \frac{1}{\cos^2 x}\cdot\frac{1}{\sin^2 x} = \sec^2 x\,\csc^2 x = \text{RHS} \quad \text{(A1)}$

Identity proved.

Where the marks are won and lost

• Pick one side and stay on it. Transforming the LHS into the RHS is a clean proof; shuffling terms across the equals sign is not, and examiners penalise it.
• The key move is the common denominator $\sin^2 x\cos^2 x$, which sets up the Pythagorean identity in the numerator.
• End by explicitly showing the result equals the RHS, so the proof visibly closes.

Common mistakes

• Treating the identity as an equation and “solving” it.
• Forgetting that $\sec^2 x = \frac{1}{\cos^2 x}$ (not $\frac{1}{\cos x}$).
• Stopping at $\frac{1}{\sin^2 x\cos^2 x}$ without showing it equals $\sec^2 x\csc^2 x$.

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