Question

If $3\cot A=4,then\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}={\cos }^{2}A-{\sin }^{2}A$
Enter $1$ for true and $0$ for false

Answer 1

$\cot A=\frac{4}{3}=\frac{b}{p}b=4,p=3h=\sqrt{4^2+3^2}=5\cos A=\frac{b}{h}=\frac{4}{5}\sin A=\frac{p}{h}=\frac{3}{5}tanA=\frac{p}{b}=\frac{3}{4}$

LHS$=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$

$\frac{1-{\left(\frac{3}{4}\right)}^2}{1+{\left(\frac{3}{4}\right)}^2}=\frac{7}{25}$

RHS$={\cos }^{2}A-{\sin }^{2}A$

$={\left(\frac{4}{5}\right)}^2-{\left(\frac{3}{5}\right)}^2=\frac{7}{25}$

Here, LHS$=$RHS