Mathematics — Standard (041) — CBSE Class 10 board question

Given: $p = \dfrac{\sec\alpha}{\csc\beta}$ and $q = \dfrac{\tan\alpha}{\csc\beta}$

LHS $= (p^2 - q^2)\sec^2\alpha$

$$= \left(\frac{\sec^2\alpha}{\csc^2\beta} - \frac{\tan^2\alpha}{\csc^2\beta}\right)\sec^2\alpha$$

$$= \frac{(\sec^2\alpha - \tan^2\alpha)}{\csc^2\beta} \cdot \sec^2\alpha$$

Using the identity $\sec^2\alpha - \tan^2\alpha = 1$:

$$= \frac{1}{\csc^2\beta} \cdot \sec^2\alpha = \frac{\sec^2\alpha}{\csc^2\beta} = p^2 = \textbf{RHS}$$

Hence proved. $\blacksquare$

Source: Chapter 8, Section 8.4 — Trigonometric Identities

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• The key identity used is $\sec^2\alpha - \tan^2\alpha = 1$ (from Section 8.4).
• Substitute $p^2$ and $q^2$ directly from the given expressions — both have the same denominator $\csc^2\beta$, so they combine neatly.
• After simplification, the result $\dfrac{\sec^2\alpha}{\csc^2\beta}$ is exactly $p^2$, completing the proof.
• Always write LHS = … = RHS and state which identity you applied — examiners award a step mark for citing $\sec^2\alpha - \tan^2\alpha = 1$.