Question : Let ABC be a right-angled triangle where $\angle \mathrm{A}=90^{\circ}$ and $\angle \mathrm{C}=45^{\circ}$. Find the value of $\sec \mathrm{C}+\sin \mathrm{C} \sec \mathrm{C}$.
Option 1: $1$
Option 2: $1-\sqrt{2}$
Option 3: $1+\sqrt{2}$
Option 4: $\sqrt{2}-1$
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Correct Answer: $1+\sqrt{2}$
Solution :
The value of $\angle C = 45°$
$\sec C = \sec 45° = \sqrt{2}$
$\sin C = \sin 45° = \frac{1}{\sqrt{2}}$
Now, $\sec \mathrm{C}+\sin \mathrm{C} \sec \mathrm{C}$
$=\sqrt{2}+ \frac{1}{\sqrt{2}} \times \sqrt{2}$
$= \sqrt{2}+1$
Hence, the correct answer is $1+\sqrt{2}$.
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