જો $3 \cot A=4$ હોય, તો નક્કી કરો કે $\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$ છે કે નહિ.

It is given that $3 \cot A=4$

Or, cot $A =\frac{4}{3}$

Consider a right triangle $ABC$, right-angled at point $B$.

$\cot A=\frac{\text { Side adjacent to } \angle A}{\text { Side opposite to } \angle A}$

$\frac{A B}{B C}=\frac{4}{3}$

If $AB$ is $4 k$, then $BC$ will be $3 k$, where $k$ is a positive integer.

$\ln \triangle ABC$

$(A C)^{2}=(A B)^{2}+(B C)^{2}$

$=(4 k)^{2}+(3 k)^{2}$

$=16 k^{2}+9 k^{2}$

$=25 k^{2}$

$A C=5 k$

$\cos A=\frac{\text { Side adjacent to } \angle A }{\text { Hypotenuse }}=\frac{ AB }{ AC }$

$=\frac{4 k}{5 k }=\frac{4}{5}$

$\sin A=\frac{\text { Side opposite to } \angle A }{\text { Hypotenuse }}=\frac{ BC }{ AC }$

$=\frac{3 k }{5 k }=\frac{3}{5}$

$\tan A=\frac{\text { Side opposite to } \angle A }{\text { Hypotenuse }}=\frac{ BC }{ AB }$

$=\frac{3 k }{4 k}=\frac{3}{4}$

$\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$

$=\frac{\frac{7}{16}}{\frac{25}{16}}=\frac{7}{25}$

$\cos ^{2} A-\sin ^{2} A=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}$

$=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$

$\quad \frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$