Mathematics — CBSE Class 10 board question

LHS $= \dfrac{\tan\theta}{1-\cot\theta} + \dfrac{\cot\theta}{1-\tan\theta}$

Convert to $\sin\theta$ and $\cos\theta$:

$$= \frac{\dfrac{\sin\theta}{\cos\theta}}{1 - \dfrac{\cos\theta}{\sin\theta}} + \frac{\dfrac{\cos\theta}{\sin\theta}}{1 - \dfrac{\sin\theta}{\cos\theta}}$$

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

$$= \frac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)} - \frac{\cos^2\theta}{\sin\theta(\sin\theta - \cos\theta)}$$

$$= \frac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta - \cos\theta)}$$

Using $a^3 - b^3 = (a-b)(a^2+ab+b^2)$:

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

$$= \frac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} + 1 = 1 + \sec\theta\,\text{cosec}\,\theta = \textbf{RHS}$$

Source: Exercise 8.3, Q.4(iii), Chapter 8 — Introduction to Trigonometry

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• Key hint: Convert everything into $\sin\theta$ and $\cos\theta$ first (the textbook hint says exactly this).
• After getting a common denominator, factor the numerator using the difference of cubes identity — this is the crucial step examiners look for.
• Use $\sin^2\theta + \cos^2\theta = 1$ to simplify the bracket to $1 + \sin\theta\cos\theta$.
• The final split $\dfrac{1}{\sin\theta\cos\theta} + 1 = \sec\theta\,\text{cosec}\,\theta + 1$ must be written explicitly to earn full marks.