Question

The value of \(tan^{-1}(\frac{cos(\frac{19π}{4})-1}{sin(\frac{π}{4})})\) is equal to

• A. \(-\frac{π}{4}\)
• B. \(-\frac{5π}{12}\)
• C. \(-\frac{3π}{8}\)
• D. \(-\frac{4π}{9}\)

Hint:

Reduce 19pi/4 to a simple angle and recognise tan(3pi/8) = 1 + sqrt 2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Reduce the angle using a period of \(2\pi\): \(\frac{19\pi}{4} = 4\pi + \frac{3\pi}{4}\).

Step 2: Evaluate the fraction
\(\cos\frac{19\pi}{4} = \cos\frac{3\pi}{4} = -\frac{1}{\sqrt2}\) and \(\sin\frac\pi4 = \frac{1}{\sqrt2}\).
\[ \frac{-\frac{1}{\sqrt2} - 1}{\frac{1}{\sqrt2}} = -1 - \sqrt2 = -(1 + \sqrt2) \]

Step 3: Inverse tangent
\(\tan\frac{3\pi}{8} = \tan 67.5^{\circ} = 1 + \sqrt2\). Because \(\tan^{-1}\) is odd and its range is \((-\frac\pi2, \frac\pi2)\):
\[ \tan^{-1}\big(-(1 + \sqrt2)\big) = -\frac{3\pi}{8} \]
Option (A), \(-\pi/4\), has tan equal to \(-1\), which is not the value.

Final Answer:
The value is \(-\frac{3\pi}{8}\), option (C). \[ \boxed{-\frac{3\pi}{8}} \]