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

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

Multiply numerator and denominator inside the root by $(1+\sin A)$:

$$= \sqrt{\dfrac{(1+\sin A)^2}{(1-\sin A)(1+\sin A)}} = \sqrt{\dfrac{(1+\sin A)^2}{1-\sin^2 A}}$$

Since $1 - \sin^2 A = \cos^2 A$:

$$= \sqrt{\dfrac{(1+\sin A)^2}{\cos^2 A}} = \dfrac{1+\sin A}{\cos A}$$

$$= \dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A} = \sec A + \tan A = \textbf{RHS}$$ $\blacksquare$

Source: Introduction to Trigonometry, Exercise 8.3 Q4(vi)

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• The standard trick is to rationalise by multiplying inside the square root by $(1+\sin A)$ in both numerator and denominator, turning the denominator into $(1-\sin^2 A) = \cos^2 A$.
• This lets you take the square root cleanly, then split the fraction into $\frac{1}{\cos A} + \frac{\sin A}{\cos A}$.
• Examiners award 1 mark for the rationalisation/identity step and 1 mark for the correct simplification to $\sec A + \tan A$. Show every step clearly.