If $\tan \theta=\frac{24}{7}$, find that $\sin \theta+\cos \theta$
Given:
$\tan \theta=\frac{24}{7}$....(1)
To find:
$\sin \theta+\cos \theta$
Now we know $\tan \theta$ is defined as follows
$\tan \theta=\frac{\text { Perpendicular side opposite to } \angle \theta}{\text { Base side adjacent to } \angle \theta}$....(2)
Now by comparing equation (1) and (2)
We get
Perpendicular side opposite to $\angle \theta=24$
Base side adjacent to $\angle \theta=7$
Therefore triangle representing angle $\theta$ is as shown below
Side AC is unknown and can be found using Pythagoras theorem
Therefore,
$A C^{2}=A B^{2}+B C^{2}$
Now by substituting the value of known sides from figure
We get,
$A C^{2}=24^{2}+7^{2}$
$=576+49$
$=625$
Now by taking square root on both sides
We get,
$A C=\sqrt{625}$
$=25$
Therefore Hypotenuse side AC = 25 …… (3)
Now we know, $\sin \theta$ is defined as follows
$\sin \theta=\frac{\text { Perpendicular side opposite to } \angle \theta}{\text { Hypotenuse }}$
Therefore from figure (a) and equation (3)
We get,
$\sin \theta=\frac{A B}{A C}$
$=\frac{24}{25}$
$\sin \theta=\frac{24}{25}$....(4)
Now we know, $\cos \theta$ is defined as follows
$\cos \theta=\frac{\text { Base side adjacent to } \angle \theta}{\text { Hypotenuse }}$
Therefore from figure (a) and equation (3)
We get,
$\cos \theta=\frac{B C}{A C}$
$=\frac{7}{25}$
$\cos \theta=\frac{7}{25}$.....(5)
Now we need to find the value of expression $\sin \theta+\cos \theta$
Therefore by substituting the value of $\sin \theta$ and $\cos \theta$ from equation (4) and (5) respectively, we get,
$\sin \theta+\cos \theta=\frac{24}{25}+\frac{7}{25}$
$=\frac{24+7}{25}$
$=\frac{31}{25}$
Hence $\sin \theta+\cos \theta=\frac{31}{25}$