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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.
Solution

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.
$\ln \triangle ABC$
$(A C)^{2}=(A B)^{2}+(B C)^{2}$
$=(4 k)^{2}+(3 k)^{2}$
$=16 k^{2}+9 k^{2}$
$=25 k^{2}$
$A C=5 k$
$\cos A=\frac{\text { Side adjacent to } \angle A }{\text { Hypotenuse }}=\frac{ AB }{ AC }$
$=\frac{4 k}{5 k }=\frac{4}{5}$
$\sin A=\frac{\text { Side opposite to } \angle A }{\text { Hypotenuse }}=\frac{ BC }{ AC }$
$=\frac{3 k }{5 k }=\frac{3}{5}$
$\tan A=\frac{\text { Side opposite to } \angle A }{\text { Hypotenuse }}=\frac{ BC }{ AB }$
$=\frac{3 k }{4 k}=\frac{3}{4}$
$\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$
$=\frac{\frac{7}{16}}{\frac{25}{16}}=\frac{7}{25}$
$\cos ^{2} A-\sin ^{2} A=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}$
$=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$
$\quad \frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$