Q4(x):

Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (x) $(\frac{1 + \tan^2 A}{1 + \cot^2 A}) = (\frac{1 - \tan A}{1 - \cot A})^2 = \tan^2 A$

Given: An identity involving trigonometric ratios of an acute angle $A$: $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$.

To Prove: 1. $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \tan^2 A$ 2. $\left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$

Step 1: Proving the first part $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \tan^2 A$

We use the fundamental trigonometric identities:
$1 + \tan^2 A = \sec^2 A$
$1 + \cot^2 A = \csc^2 A$
[Since $\sec^2 \theta - \tan^2 \theta = 1$ and $\csc^2 \theta - \cot^2 \theta = 1$]

Substitute these into the expression:
$\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A}$

Express $\sec A$ and $\csc A$ in terms of $\sin A$ and $\cos A$:
$\sec A = \frac{1}{\cos A}$ and $\csc A = \frac{1}{\sin A}$

Therefore:
$\frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1} = \frac{\sin^2 A}{\cos^2 A}$

Using the identity $\tan A = \frac{\sin A}{\cos A}$:
$\frac{\sin^2 A}{\cos^2 A} = \tan^2 A$

Thus, the first part is proved: $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \tan^2 A$.

Step 2: Proving the second part $\left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$

Express $\cot A$ in terms of $\tan A$:
$\cot A = \frac{1}{\tan A}$

Substitute this into the expression:
$\left(\frac{1 - \tan A}{1 - \frac{1}{\tan A}}\right)^2 = \left(\frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}}\right)^2$

Simplify the fraction:
$\left((1 - \tan A) \times \frac{\tan A}{\tan A - 1}\right)^2$

Note that $(1 - \tan A) = -( \tan A - 1)$:
$\left(\frac{-( \tan A - 1) \times \tan A}{\tan A - 1}\right)^2 = (-\tan A)^2$

Calculate the square:
$(-\tan A)^2 = \tan^2 A$

Conclusion:
Since both the first expression and the second expression are equal to $\tan^2 A$, the identity is proved:
$\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$

Final Answer: The identity is proven as both sides simplify to $\tan^2 A$.