Mathematics — CBSE Class 10 board question

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

Taking LCM:

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

$$= \tan A \cdot \frac{-2\sec A}{1-\sec^2 A}$$

Since $\sec^2 A - \tan^2 A = 1 \Rightarrow 1 - \sec^2 A = -\tan^2 A$:

$$= \tan A \cdot \frac{-2\sec A}{-\tan^2 A}$$

$$= \frac{2\sec A}{\tan A} = \frac{2 \cdot \dfrac{1}{\cos A}}{\dfrac{\sin A}{\cos A}} = \frac{2}{\sin A} = 2\,\cosec A = \textbf{RHS}$$

Hence proved.

Source: Exercise 8.3, Chapter 8

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• The key step is combining the two fractions over a common denominator $(1+\sec A)(1-\sec A) = 1-\sec^2 A$.
• Then substitute $1 - \sec^2 A = -\tan^2 A$ (from the identity $\sec^2 A - \tan^2 A = 1$).
• Finally, simplify $\dfrac{\sec A}{\tan A} = \dfrac{1/\cos A}{\sin A/\cos A} = \dfrac{1}{\sin A} = \cosec A$.
• Examiners award marks for: correct LCM step, correct identity use, and final simplification. Show each step clearly.