Mathematics — CBSE Class 10 board question

Divide numerator and denominator of LHS by $\cos\theta$:

$$\text{LHS} = \frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac{\tan\theta - 1 + \sec\theta}{\tan\theta + 1 - \sec\theta}$$

$$= \frac{(\tan\theta + \sec\theta) - 1}{(\tan\theta - \sec\theta) + 1}$$

Multiply numerator and denominator by $(\tan\theta - \sec\theta)$:

$$= \frac{[(\tan\theta + \sec\theta) - 1](\tan\theta - \sec\theta)}{[(\tan\theta - \sec\theta) + 1](\tan\theta - \sec\theta)}$$

$$= \frac{(\tan^2\theta - \sec^2\theta) - (\tan\theta - \sec\theta)}{(\tan\theta - \sec\theta + 1)(\tan\theta - \sec\theta)}$$

Using $\sec^2\theta - \tan^2\theta = 1$, so $\tan^2\theta - \sec^2\theta = -1$:

$$= \frac{-1 - (\tan\theta - \sec\theta)}{(\tan\theta - \sec\theta + 1)(\tan\theta - \sec\theta)}$$

$$= \frac{-(1 + \tan\theta - \sec\theta)}{(\tan\theta - \sec\theta + 1)(\tan\theta - \sec\theta)} = \frac{-1}{\tan\theta - \sec\theta} = \frac{1}{\sec\theta - \tan\theta} = \text{RHS}$$

Hence proved. $\blacksquare$

Source: Chapter 8, Example 12 (8.4 Trigonometric Identities)

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• Key trick: Divide throughout by $\cos\theta$ to convert sin/cos into tan and sec, then multiply by $(\tan\theta - \sec\theta)$ to use the identity $\sec^2\theta - \tan^2\theta = 1$.
• Examiners award 1 mark for the division step, 1 mark for correct multiplication and use of the identity, and 1 mark for reaching the RHS cleanly.
• Always state which identity you are using (here: $\sec^2\theta = 1 + \tan^2\theta$) — it secures method marks.