Question

If $3\cot A=4$ check whether $\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}={\cos }^{2}A-{\sin }^{2}A$ or not

Answer 1

$cotA=\left(\frac{4}{3}\right)\therefore \left(\frac{3}{4}\right)LHS\left(\frac{1-ta{n}^{2}A}{1+ta{n}^{2}A}\right)=\left(\frac{1-\left(\right(\frac{3}{4}{)}^2)}{1+\left(\right(\frac{3}{4}{)}^2)}\right)=\left(\frac{1-\left(\right(\frac{9}{16}\left)\right)}{1+\left(\right(\frac{9}{16}\left)\right)}\right)=\left(\frac{7}{25}\right)astanA=\left(\frac{3}{4}\right)socosA=\left(\frac{4}{5}\right)andsinA=\left(\frac{3}{5}\right)RHS=co{s}^{2}A-si{n}^{2}A=\left(\right(\frac{4}{5}{)}^2-\left(\frac{3}{5}{)}^2\right)=\left(\frac{16}{25}\right)-\left(\frac{9}{25}\right)=\left(\frac{7}{25}\right)\therefore LHS=RHS$