MCQ
If $\tan\theta=\frac{4}{3}$ then $(\sin\theta+\cos\theta)=?$
- A$\frac{7}{3}$
- B$\frac{7}{4}$
- ✓$\frac{7}{5}$
- D$\frac{5}{7}$
✓
Answer
Correct option: C.
$\frac{7}{5}$
Consider $\triangle\text{ABC}$ where $\angle\text{A}=90^\circ,\angle\text{A}=\theta$
Then, $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{BC}}{\text{AB}}=\frac{4}{3}$
Let $BC = 4k$ and $AB = 3k,$ where $k$ is positive.
By pythagoras theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(\text{3k})^2+(\text{4k})^2$
$=\text{9k}^2+\text{16k}^2=\text{25k}^2$
$\Rightarrow\text{AC}=\text{5k}$
Now, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\text{4k}}{\text{5k}}=\frac{4}{5}$
And, $\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{3\text{k}}{\text{5k}}=\frac{3}{5}$
$\therefore\sin\theta+\cos\theta=\frac{4}{5}+\frac{3}{5}=\frac{7}{5}$
Then, $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{BC}}{\text{AB}}=\frac{4}{3}$
Let $BC = 4k$ and $AB = 3k,$ where $k$ is positive.
By pythagoras theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(\text{3k})^2+(\text{4k})^2$
$=\text{9k}^2+\text{16k}^2=\text{25k}^2$
$\Rightarrow\text{AC}=\text{5k}$
Now, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\text{4k}}{\text{5k}}=\frac{4}{5}$
And, $\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{3\text{k}}{\text{5k}}=\frac{3}{5}$
$\therefore\sin\theta+\cos\theta=\frac{4}{5}+\frac{3}{5}=\frac{7}{5}$
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