Question

If $4\cot A=3$ then find the value of $cose{c}^{2}A-{\cot }^{2}A$.

Answer 1

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

Thus $\cot A=\frac{3}{4}=\frac{B}{P}$

$P=4$, $B=3$

Now, $H^2=P^2+B^2$

$H^2=3^2+4^2$

$H^2=5^2$

$H=5$

$cose{c}^{2}A-co{t}^{2}A={\left(\frac{H}{P}\right)}^2-{\left(\frac{B}{P}\right)}^2$

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

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

$=$ $\frac{16}{16}$

$=$ $1$