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

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

Convert to $\sin A$ and $\cos A$: let $s = \sin A$, $c = \cos A$.

$$= \frac{\tfrac{s}{c}}{1-\tfrac{c}{s}} + \frac{\tfrac{c}{s}}{1-\tfrac{s}{c}} = \frac{\tfrac{s}{c}}{\tfrac{s-c}{s}} + \frac{\tfrac{c}{s}}{\tfrac{c-s}{c}}$$

$$= \frac{s^2}{c(s-c)} + \frac{c^2}{s(c-s)} = \frac{s^2}{c(s-c)} - \frac{c^2}{s(s-c)}$$

$$= \frac{1}{s-c}\left(\frac{s^2}{c} - \frac{c^2}{s}\right) = \frac{1}{s-c}\cdot\frac{s^3 - c^3}{sc}$$

$$= \frac{(s-c)(s^2+sc+c^2)}{sc(s-c)} = \frac{s^2+c^2+sc}{sc}$$

$$= \frac{1 + sc}{sc} \quad (\because s^2+c^2=1)$$

$$= \frac{1}{sc} + 1 = 1 + \sec A\,\text{cosec}\,A = \textbf{RHS}$$

Hence proved. $\blacksquare$

Source: Exercise 8.3, Q4(iii), Chapter 8

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• The key hint (given in textbook) is to convert everything to $\sin A$ and $\cos A$.
• The algebraic trick is recognising $s^3 - c^3 = (s-c)(s^2+sc+c^2)$, which cancels $(s-c)$ in the denominator.
• Then use the identity $\sin^2 A + \cos^2 A = 1$ to simplify the numerator.
• Finally, $\dfrac{1}{\sin A \cos A} = \text{sec}\,A\,\text{cosec}\,A$.
• Examiners award marks for: correct conversion (1 mark), correct algebraic simplification (1 mark), and using $\sin^2A+\cos^2A=1$ to reach RHS (1 mark).