Question

If \(\sin x = p\), then prove that : (i) \(\cot x = \frac{\sqrt{1 - p^2}}{p}\) (ii) \(\frac{1 + \tan^2 x}{1 + \cot^2 x} = \frac{p^2}{1 - p^2}\)

Hint:

The expression \(\frac{1 + \tan^2 \theta}{1 + \cot^2 \theta}\) always simplifies to \(\tan^2 \theta\). Memorizing this can save you significant time in exams.

Answer 1

Step 1: Solving OR (B):
1. Given \(\sin x = p\), then \(\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - p^2}\).
2. Part (i): \(\cot x = \frac{\cos x}{\sin x} = \frac{\sqrt{1 - p^2}}{p}\). (Proved)

3. Part (ii): We know \(\frac{1 + \tan^2 x}{1 + \cot^2 x} = \frac{\sec^2 x}{\csc^2 x} = \frac{1/\cos^2 x}{1/\sin^2 x} = \tan^2 x\).
4. Since \(\tan x = \frac{\sin x}{\cos x} = \frac{p}{\sqrt{1 - p^2}}\), then \(\tan^2 x = \frac{p^2}{1 - p^2}\). (Proved)
Step 2: Final Answer (OR):
Both identities are proved based on \(\sin x = p\).