Question

Given : $4\sin \theta =3\cos \theta$, find the value of:

$cose{c}^2\theta -{\cot }^2\theta .$

Answer 1

$4\sin \Theta =3\cos \Theta$

$\tan \Theta =\frac{3}{4}$

$\tan \Theta =\frac{P}{B}=\frac{3}{4}$

$P=3,B=4$

Using Pythagoras Theorem,

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

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

$H=5$

${\cot }^2\Theta -cose{c}^2\Theta =(\frac{B}{P}{)}^2-(\frac{H}{P}{)}^2$

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

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

= $-1$