Question

Evaluate \[ \frac{d}{dx}\left(\sin^{-1}(\tan x)+\tan^{-1}(\sin x)\right). \]

• A. \[ \frac{1}{\cos x\sqrt{\cos2x}} +\frac{2\cos x}{3-\cos2x} \]
• B. \[ \frac{\sec x}{\sqrt{1+\tan^2x}} +\frac{\cos x}{1-\sin^2x} \]
• C. \[ \frac{\sec x\tan x}{\sqrt{\cos2x}} +\frac{\sin x}{1+\cos2x} \]
• D. \[ \frac{\sec x}{1-\tan^2x} +\frac{\cos x}{\sqrt{1+\sin^2x}} \]

Hint:

While differentiating inverse trigonometric functions, first apply the chain rule and then simplify using identities such as \[ \boxed{1-\tan^2x=\frac{\cos2x}{\cos^2x}} \] and \[ \boxed{1+\sin^2x=\frac{3-\cos2x}{2}.} \]

Answer 1

The Correct Option is A

Step 1: Differentiate the first term. Using \[ \frac{d}{dx}\left(\sin^{-1}u\right) = \frac{u'}{\sqrt{1-u^2}}, \] with \[ u=\tan x, \] we get \[ \frac{d}{dx}\left(\sin^{-1}(\tan x)\right) = \frac{\sec^2x}{\sqrt{1-\tan^2x}}. \] Now, \[ 1-\tan^2x = \frac{\cos2x}{\cos^2x}. \] Hence, \[ \frac{\sec^2x}{\sqrt{1-\tan^2x}} = \frac{1}{\cos x\sqrt{\cos2x}}. \]

Step 2: Differentiate the second term. Using \[ \frac{d}{dx}\left(\tan^{-1}u\right) = \frac{u'}{1+u^2}, \] with \[ u=\sin x, \] we obtain \[ \frac{\cos x}{1+\sin^2x}. \] Now, \[ 1+\sin^2x = 1+\frac{1-\cos2x}{2} = \frac{3-\cos2x}{2}. \] Therefore, \[ \frac{\cos x}{1+\sin^2x} = \frac{2\cos x}{3-\cos2x}. \]

Step 3: Add the derivatives. Hence, \[ \frac{d}{dx} \left( \sin^{-1}(\tan x) +\tan^{-1}(\sin x) \right) = \frac{1}{\cos x\sqrt{\cos2x}} + \frac{2\cos x}{3-\cos2x}. \] Therefore, \[ \boxed{ \frac{1}{\cos x\sqrt{\cos2x}} +\frac{2\cos x}{3-\cos2x} }. \] Thus, \[ \boxed{(A)} \] is the correct answer.