Question

If \( 3\sin^2 x + 10\cos x - 6 = 0 \), \( 0^\circ<x<90^\circ \), then the value of \( \sec x + \cosec x + \cot x \) is:

• A. \( 4 - \sqrt{3} \)
• B. \( 2 + \sqrt{3} \)
• C. \( 3 - \sqrt{2} \)
• D. \( 3 + \sqrt{2} \)

Hint:

Reduce everything to one trig function (cos or sin) to simplify. Always reject roots outside \( (0, 1) \) for cosine or sine in \( 0^\circ<x<90^\circ \).

Answer 1

The Correct Option is D

Use identity: \[ \sin^2 x = 1 - \cos^2 x \] Substitute in equation: \[ 3(1 - \cos^2 x) + 10 \cos x - 6 = 0 \Rightarrow 3 - 3 \cos^2 x + 10 \cos x - 6 = 0
-3 \cos^2 x + 10 \cos x - 3 = 0 \Rightarrow 3 \cos^2 x - 10 \cos x + 3 = 0 \] Solve quadratic: \[ \cos x = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6} \Rightarrow \cos x = 3 \text{ (reject)},\ \cos x = \frac{1}{3} \] Now find: \[ \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \] Then: \[ \sec x = \frac{1}{\cos x} = 3,\quad \csc x = \frac{1}{\sin x} = \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4},\quad \cot x = \frac{\cos x}{\sin x} = \frac{1/3}{2\sqrt{2}/3} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} \] Add: \[ \sec x + \csc x + \cot x = 3 + \frac{3\sqrt{2}}{4} + \frac{\sqrt{2}}{4} = 3 + \frac{4\sqrt{2}}{4} = 3 + \sqrt{2} \]