Question

If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin² A or not

Answer 1

Given 3cot A = 4, it follows that cot A = \( \frac{4}{3} \). Consider a right triangle ABC, with the right angle at B.

As cot(A) = \( \frac{\text{adjacent}}{\text{opposite}} = \frac{AB}{BC} = \frac{4}{3} \), let AB = 4k and BC = 3k, where k is a positive integer.

In \( \Delta \)ABC, by the Pythagorean theorem:

(AC)² = (AB)² + (BC)²

= (4k)² + (3k)²

= 16k² + 9k²

= 25k²

Therefore, AC = 5k.

From these lengths, we find:

tan(A) = \( \frac{BC}{AB} = \frac{3}{4} \)

sin(A) = \( \frac{BC}{AC} = \frac{3}{5} \)

cos(A) = \( \frac{AB}{AC} = \frac{4}{5} \)

Now, let's evaluate the expression \( \frac{1-\tan^2A}{1+\tan^2A} \):

\(\frac{1-\tan^2A}{1+\tan^2A}=\frac{1-(\frac{3}{4})^2}{1+(\frac{3}{4})^2}\)

\(= \frac{1-\frac{9}{16}}{1+\frac{9}{16}}\)

\(= \frac{\frac{16-9}{16}}{\frac{16+9}{16}}\)

\(= \frac{\frac{7}{16}}{\frac{25}{16}}\)

\(= \frac{7}{25}\)

Next, let's evaluate cos² A - sin² A:

cos² A - sin² A = \( (\frac{4}{5})^2 - (\frac{3}{5})^2 \)

\(= \frac{16}{25} - \frac{9}{25}\)

\(= \frac{16-9}{25}\)

\(= \frac{7}{25}\)

Thus, it is verified that \( \frac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A \).