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

It is given that 3cot A = 4 
Or, cot A =\(\frac{4}{3}\)
Consider a right triangle ABC, right-angled at point B.

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

If AB is 4k, then BC will be 3k, where k is a positive integer. 
In \(Δ\)ABC,
(AC)² = (AB)² + (BC)²
= (4k)² + (3k)²
= 16k² + 9k²
= 25k² 
\(⇒\) AC = 5k

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

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

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

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

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

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

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

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

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

\(∴ \frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)} = \text{cos}^2 A –\text{ sin }^2 A\)