Question:
If $\tan A=\frac{5}{12}$, find the value of $(\sin \mathrm{A}+\cos \mathrm{A}) \sec \mathrm{A} .$
Solution:
Given: $\tan A=\frac{5}{12}$
$\frac{\text { Perpendicular }}{\text { Base }}=\frac{5}{12}$
Perpendicular $=5$
Base $=12$
Hypotenuse $=\sqrt{(\text { Perpendicular })^{2}+(\text { Base })^{2}}$
We know that: $\tan A=\frac{\text { perpendicular }}{\text { Base }}$
Hypotenuse $=\sqrt{(5)^{2}+(12)^{2}}$
Hypotenuse $=\sqrt{169}$
Hypotenuse $=13$
Now we find, $(\sin A+\cos A) \sec A$
$\Rightarrow(\sin A+\cos A) \sec A=\left(\frac{5}{13}+\frac{12}{13}\right) \times \frac{13}{12}$
$\Rightarrow(\sin A+\cos A) \sec A=\frac{17}{13} \times \frac{13}{12}$
$\Rightarrow(\sin A+\cos A) \sec A=\frac{17}{12}$
Hence the value of $(\sin A+\cos A) \sec A$ is $\frac{17}{12}$