Question
If $\cos\theta=0.6,$ show that $(5\sin\theta-3\tan\theta)=0.$
✓
Answer
Given: $\cos\theta=0.6=\frac{6}{10}=\frac{3}{5}$
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
Then, $\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac35$
Let $AB = 3k$
and $AC = 5k$
Where k is positive
By Pythagoras theorem, we have
$(AC)^2 = (AB)^2 + (BC)^2$
$\Rightarrow (BC)^2 = (AC)^2 - (AB)^2$
$= [(5k)^2 - (3k)^2] = 16k^2$
$= 16k^2$
$\Rightarrow BC = 4k$
$\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{4\text{k}}{5\text{k}}=\frac{4}{5}$
$\cos\theta=\frac{\sin\theta}{\cos\theta}=\Big(\frac45\times\frac53\Big)=\frac{4}{3}$
$\Rightarrow(5\sin\theta-3\tan\theta)=\Big(5\times\frac{4}{5}-3\times\frac43\Big)=0$
Hence, $(5\sin\theta-3\tan\theta)=0$
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
Then, $\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac35$
Let $AB = 3k$
and $AC = 5k$
Where k is positive
By Pythagoras theorem, we have
$(AC)^2 = (AB)^2 + (BC)^2$
$\Rightarrow (BC)^2 = (AC)^2 - (AB)^2$
$= [(5k)^2 - (3k)^2] = 16k^2$
$= 16k^2$
$\Rightarrow BC = 4k$
$\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{4\text{k}}{5\text{k}}=\frac{4}{5}$
$\cos\theta=\frac{\sin\theta}{\cos\theta}=\Big(\frac45\times\frac53\Big)=\frac{4}{3}$
$\Rightarrow(5\sin\theta-3\tan\theta)=\Big(5\times\frac{4}{5}-3\times\frac43\Big)=0$
Hence, $(5\sin\theta-3\tan\theta)=0$
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