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

Concept Used:
We use fundamental trigonometric identities:
1) $\sin^2 x + \cos^2 x = 1$
2) $\tan x = \frac{\sin x}{\cos x}$
3) $\cot x = \frac{\cos x}{\sin x}$
4) $1 + \tan^2 x = \sec^2 x$ and $1 + \cot^2 x = \csc^2 x$

(i) Proof of $\cot x = \frac{\sqrt{1 - p^2}}{p}$:
Given $\sin x = p$

Using identity:
$\sin^2 x + \cos^2 x = 1$
$p^2 + \cos^2 x = 1$
$\cos^2 x = 1 - p^2$
$\cos x = \sqrt{1 - p^2}$ (taking positive value for acute angle)

Now,
$\cot x = \frac{\cos x}{\sin x}$
$= \frac{\sqrt{1 - p^2}}{p}$

Hence proved.

(ii) Proof of $\frac{1 + \tan^2 x}{1 + \cot^2 x} = \frac{p^2}{1 - p^2}$:
Using identities:
$1 + \tan^2 x = \sec^2 x$
$1 + \cot^2 x = \csc^2 x$

Therefore,
\[ \frac{1 + \tan^2 x}{1 + \cot^2 x} = \frac{\sec^2 x}{\csc^2 x} \]
Now substitute values:
$\sec^2 x = \frac{1}{\cos^2 x}$ and $\csc^2 x = \frac{1}{\sin^2 x}$

\[ = \frac{\frac{1}{\cos^2 x}}{\frac{1}{\sin^2 x}} = \frac{\sin^2 x}{\cos^2 x} \]
But $\sin x = p$ and $\cos^2 x = 1 - p^2$

\[ = \frac{p^2}{1 - p^2} \]
Hence proved.

Final Result:
(i) $\cot x = \frac{\sqrt{1 - p^2}}{p}$
(ii) $\frac{1 + \tan^2 x}{1 + \cot^2 x} = \frac{p^2}{1 - p^2}$