Step 1: Simplify the expression inside \( \tan^{-1} \)
The given expression is: \[ \tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right). \] Using trigonometric identities, rewrite: \[ 1 - \sin x = (\cos^2\frac{x}{2} + \sin^2\frac{x}{2}) - 2\sin\frac{x}{2}\cos\frac{x}{2} = (\cos\frac{x}{2} - \sin\frac{x}{2})^2. \] Step 2: Transform into a single tangent function
Substituting \( 1 - \sin x \) and \( \cos x = (\cos^2\frac{x}{2} - \sin^2\frac{x}{2}) \): \[ \tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right) = \tan^{-1} \left[ \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \right]. \] Step 3: Simplify using \( \tan^{-1} \tan y = y \)
Since \( -\frac{\pi}{2}<x<\frac{\pi}{2} \), we simplify: \[ \tan^{-1} \left[ \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \right] = \frac{\pi}{4} + \frac{x}{2}. \] Conclusion: The simplest form is \( \frac{\pi}{4} + \frac{x}{2} \).
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
The maximum value of \( Z = 4x + y \) for a L.P.P. whose feasible region is given below is: 