Mathematics — CBSE Class 10 board question

LHS $= \dfrac{\cos A + \sin A - 1}{\cos A - \sin A + 1}$

Divide numerator and denominator by $\sin A$:

$$= \frac{\cot A + 1 - \text{cosec } A}{\cot A - 1 + \text{cosec } A}$$

$$= \frac{(\cot A - \text{cosec } A) + 1}{(\cot A + \text{cosec } A) - 1}$$

Use identity: $\text{cosec}^2 A - \cot^2 A = 1$, so replace $1 = \text{cosec}^2 A - \cot^2 A = (\text{cosec } A + \cot A)(\text{cosec } A - \cot A)$ in the numerator:

$$= \frac{(\cot A - \text{cosec } A) + (\text{cosec } A + \cot A)(\text{cosec } A - \cot A)}{(\cot A + \text{cosec } A) - 1}$$

$$= \frac{(\text{cosec } A - \cot A)\bigl[-1 + (\text{cosec } A + \cot A)\bigr]}{(\text{cosec } A + \cot A) - 1}$$

$$= \frac{(\text{cosec } A - \cot A)\bigl[(\text{cosec } A + \cot A) - 1\bigr]}{(\text{cosec } A + \cot A) - 1}$$

$$= \text{cosec } A - \cot A \quad = \textbf{ RHS}$$

Hence proved.

Source: Exercise 8.3, Q.4(v) — Chapter 8

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• The key trick is dividing throughout by sin A to introduce cosec and cot, matching the RHS.
• Then substitute $1 = \text{cosec}^2 A - \cot^2 A$ (given identity) in the numerator to factorise it so the common bracket $(\text{cosec } A + \cot A - 1)$ cancels.
• Note: the textbook states the companion identity $\dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A$; this question is the reciprocal form — same method applies.
• Always write LHS = … = RHS and end with "Hence proved."