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 $=(\frac{4}{5})^2-(\frac{3}{5})^2$

$=\frac{16}{25}-\frac{9}{25}$

$=\frac{7}{25}$

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