Question

ABC is a right-angled triangle with right angle at B. If \(AB = 2\) and \(AC = 5\), then \(\tan C + \sec C =\)

• A. \(\sqrt{\frac{7}{2}}\)
• B. \(\sqrt{\frac{7}{3}}\)
• C. \(\sqrt{21}\)
• D. \(2\sqrt{21}\)

Hint:

In right triangles, always clearly label opposite, adjacent, and hypotenuse before computing trig ratios.

Answer 1

The Correct Option is C

Step 1: Find missing side using Pythagoras theorem. \[ AC^2 = AB^2 + BC^2 \] \[ 25 = 4 + BC^2 \] \[ BC^2 = 21 \] \[ BC = \sqrt{21} \]

Step 2: Find \(\tan C\) and \(\sec C\). For angle \(C\): Opposite = \(AB=2\), Adjacent = \(BC=\sqrt{21}\), Hypotenuse = \(5\) \[ \tan C = \frac{2}{\sqrt{21}}, \quad \sec C = \frac{5}{\sqrt{21}} \]

Step 3: Add them. \[ \tan C + \sec C = \frac{2+5}{\sqrt{21}} \] \[ =\frac{7}{\sqrt{21}} = \sqrt{21} \] \[ \boxed{\sqrt{21}} \]