Question
If $\tan\theta=\frac{4}{3},$ show that $(\sin\theta+\cos\theta)=\frac75.$
✓
Answer
Given: $\tan\theta=\frac{\text{BC}}{\text{AB}}=\frac{4}{3}$
Let BC = 4k and AB = 3k
Where k is positive
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{BAC}=\theta$
By Pythagoras theorem, we get
$\big(\text{AC}\big)^2=\big(\text{AB}\big)^2+\big(\text{BC}\big)^2$
$\Rightarrow\big(\text{AC}\big)^2=\Big[\big(3\text{k}\big)^2+\big(4\text{k}\big)^2\Big]$
$\Rightarrow\big(\text{AC}\big)^2=\big(9\text{k}^2+16\text{k}^2\big)=25\text{k}^2$
$\therefore\text{AC}=\sqrt{9\text{k}^2}=5\text{k}$
$\sin\theta=\frac{4\text{k}}{5\text{k}}=\frac45$
$\cos\theta=\frac{3\text{k}}{5\text{k}}=\frac{3}{5}$
$\Rightarrow(\sin\theta+\cos\theta)=\Big(\frac{4}{5}+\frac35\Big)=\frac75$
Hence, $(\sin\theta+\cos\theta)=\frac75.$
Let BC = 4k and AB = 3k
Where k is positive
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{BAC}=\theta$
By Pythagoras theorem, we get
$\big(\text{AC}\big)^2=\big(\text{AB}\big)^2+\big(\text{BC}\big)^2$
$\Rightarrow\big(\text{AC}\big)^2=\Big[\big(3\text{k}\big)^2+\big(4\text{k}\big)^2\Big]$
$\Rightarrow\big(\text{AC}\big)^2=\big(9\text{k}^2+16\text{k}^2\big)=25\text{k}^2$
$\therefore\text{AC}=\sqrt{9\text{k}^2}=5\text{k}$
$\sin\theta=\frac{4\text{k}}{5\text{k}}=\frac45$
$\cos\theta=\frac{3\text{k}}{5\text{k}}=\frac{3}{5}$
$\Rightarrow(\sin\theta+\cos\theta)=\Big(\frac{4}{5}+\frac35\Big)=\frac75$
Hence, $(\sin\theta+\cos\theta)=\frac75.$
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