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

LHS $= \sec A(1 - \sin A)(\sec A + \tan A)$

Substituting $\sec A = \dfrac{1}{\cos A}$ and $\tan A = \dfrac{\sin A}{\cos A}$:

$$= \frac{1}{\cos A}(1 - \sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)$$

$$= \frac{1}{\cos A}(1 - \sin A)\cdot\frac{(1 + \sin A)}{\cos A}$$

$$= \frac{(1 - \sin A)(1 + \sin A)}{\cos^2 A}$$

$$= \frac{1 - \sin^2 A}{\cos^2 A}$$

Since $\sin^2 A + \cos^2 A = 1 \Rightarrow 1 - \sin^2 A = \cos^2 A$:

$$= \frac{\cos^2 A}{\cos^2 A} = 1 = \textbf{RHS}$$

Hence proved. $\blacksquare$

Source: Chapter 8, Section 8.3 (Trigonometric Identities)

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• Key strategy: Convert all ratios to $\sin A$ and $\cos A$ — the standard approach for proving identities.
• Critical identity used: $1 - \sin^2 A = \cos^2 A$ (derived from $\sin^2 A + \cos^2 A = 1$).
• Examiner expects: Clear step-by-step working starting from LHS, arriving at RHS. Do not manipulate both sides simultaneously.
• Note how combining $(\sec A + \tan A) = \dfrac{1+\sin A}{\cos A}$ with $(1-\sin A)$ creates the difference-of-squares pattern — recognising this quickly is the key skill.