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If tan A=512, find the value of (sin A + cos A) sec A.


If $\tan A=\frac{5}{12}$, find the value of $(\sin \mathrm{A}+\cos \mathrm{A}) \sec \mathrm{A} .$


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 $\frac{17}{12}$

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