Question

If cosecθ + cotθ = 5, then the value of tanθ is equal to

• A. 13/24
• B. 5/12
• C. 7/12
• D. 1/12
• E. 3/12

Answer 1

The Correct Option is B

Given equation:

\(\csc \theta + \cot \theta = 5\) 

We use the identity:

\(\csc \theta = \frac{1}{\sin \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}\)

Rewriting the equation:

\(\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = 5\)

Taking common denominator:

\(\frac{1 + \cos \theta}{\sin \theta} = 5\)

Rearranging:

\(1 + \cos \theta = 5 \sin \theta\)

Squaring both sides:

\((1 + \cos \theta)^2 = (5 \sin \theta)^2\)

Expanding:

\(1 + 2\cos \theta + \cos^2 \theta = 25 \sin^2 \theta\)

Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):

\(1 + 2\cos \theta + \cos^2 \theta = 25(1 - \cos^2 \theta)\)

Expanding:

\(1 + 2\cos \theta + \cos^2 \theta = 25 - 25\cos^2 \theta\)

Rearranging:

\(1 + 2\cos \theta + \cos^2 \theta + 25\cos^2 \theta = 25\)

\(26\cos^2 \theta + 2\cos \theta - 24 = 0\)

Solving the quadratic equation \(26x^2 + 2x - 24 = 0\) using the quadratic formula:

\(x = \frac{-2 \pm \sqrt{(2)^2 - 4(26)(-24)}}{2(26)}\)

\(x = \frac{-2 \pm \sqrt{4 + 2496}}{52}\)

\(x = \frac{-2 \pm \sqrt{2500}}{52}\)

\(x = \frac{-2 \pm 50}{52}\)

Solving for \(x\):

\(x = \frac{-2 + 50}{52} = \frac{48}{52} = \frac{12}{13}\)

\(x = \frac{-2 - 50}{52} = \frac{-52}{52} = -1\) (not valid as \(\cos \theta\) must be in range \([-1,1]\))

Thus, \(\cos \theta = \frac{12}{13}\).

Finding \(\sin \theta\) using \(\sin^2 \theta + \cos^2 \theta = 1\):

\(\sin^2 \theta = 1 - \left(\frac{12}{13}\right)^2\)

\(\sin^2 \theta = 1 - \frac{144}{169}\)

\(\sin^2 \theta = \frac{169 - 144}{169} = \frac{25}{169}\)

\(\sin \theta = \frac{5}{13}\) (as \(\theta\) is in the first quadrant, \(\sin \theta\) is positive)

Finding \(\tan \theta\):

\(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{5/13}{12/13} = \frac{5}{12}\)

Thus, the correct answer is:

\(\frac{5}{12}\)