Question

$3sin\alpha -4cos\alpha =0$, then the values of tan $\alpha$ is $tan\alpha =\frac{4}{3},$
if true then enter $1$ and if false then write $0$

Answer 1

$3sin\alpha -4cos\alpha =0$
$3sin\alpha =4cos\alpha$
$tan\alpha =\frac{4}{3}$
$tan\theta =\frac{P}{B}=\frac{4}{3}$
$B=3$, $P=4$
Thus, by Pythagoras Theorem,
$H^2=B^2+P^2$
$H^2=3^2+4^2$
$H^2=25$
$H=5$ cm
Since, $\theta$ is in first quadrant, all values will be positive,
$sec\alpha =\frac{H}{B}=\frac{5}{3}$
$cosec\alpha =\frac{H}{P}=\frac{5}{4}$