Simplify using half-angle substitution. Classic CBSE problem.

Write tan⁻¹(cos x/(1 + sin x)) in Simplest Form

Question

Simplify $\tan^{-1}\left(\dfrac{\cos x}{1 + \sin x}\right)$ and express it in its simplest form.

Solution — Step by Step

The denominator $1 + \sin x$ screams half-angle. We rewrite both $\cos x$ and $1 + \sin x$ using $x/2$ substitutions so something cancels.

\cos x = \cos^2\!\tfrac{x}{2} - \sin^2\!\tfrac{x}{2}

1 + \sin x = \sin^2\!\tfrac{x}{2} + \cos^2\!\tfrac{x}{2} + 2\sin\tfrac{x}{2}\cos\tfrac{x}{2} = \left(\cos\tfrac{x}{2} + \sin\tfrac{x}{2}\right)^2

Substitute back into the fraction:

\frac{\cos x}{1+\sin x} = \frac{\left(\cos\tfrac{x}{2} - \sin\tfrac{x}{2}\right)\left(\cos\tfrac{x}{2} + \sin\tfrac{x}{2}\right)}{\left(\cos\tfrac{x}{2} + \sin\tfrac{x}{2}\right)^2}

The $\left(\cos\tfrac{x}{2} + \sin\tfrac{x}{2}\right)$ cancels cleanly:

= \frac{\cos\tfrac{x}{2} - \sin\tfrac{x}{2}}{\cos\tfrac{x}{2} + \sin\tfrac{x}{2}}

Divide numerator and denominator by $\cos\tfrac{x}{2}$ :

= \frac{1 - \tan\tfrac{x}{2}}{1 + \tan\tfrac{x}{2}}

Now recall the tan subtraction identity: $\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$ .

Put $A = \pi/4$ (so $\tan A = 1$ ) and $B = x/2$ :

\tan\!\left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{1 - \tan\tfrac{x}{2}}{1 + \tan\tfrac{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)

Since $\tan^{-1}(\tan\theta) = \theta$ when $\theta$ lies in $\left(-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right)$ , and the domain here is $x \in \left(-\dfrac{\pi}{2},\, \dfrac{\pi}{2}\right)$ , the angle $\dfrac{\pi}{4} - \dfrac{x}{2}$ comfortably stays in that range.

\boxed{\dfrac{\pi}{4} - \dfrac{x}{2}}

Why This Works

The entire trick rests on recognising that $1 + \sin x$ is a perfect square in disguise. Once we write $1 + \sin x = \left(\cos\tfrac{x}{2} + \sin\tfrac{x}{2}\right)^2$ , the numerator $\cos x$ factors as a difference of squares and one factor cancels immediately.

After the cancellation, we have a ratio $\dfrac{1 - \tan\theta}{1 + \tan\theta}$ where $\theta = x/2$ . This specific form is the fingerprint of $\tan(\pi/4 - \theta)$ . Memorising this pattern saves you from repeating all steps — you will see it again in JEE Main.

The domain condition matters more than students realise. The identity $\tan^{-1}(\tan\theta) = \theta$ only holds when $\theta \in (-\pi/2, \pi/2)$ . For $x$ outside the interval $(-\pi/2, 3\pi/2)$ , the answer would need adjustment.

Alternative Method

Write $\sin x = \cos\!\left(\dfrac{\pi}{2} - x\right)$ and $\cos x = \sin\!\left(\dfrac{\pi}{2} - x\right)$ . Let $\phi = \dfrac{\pi}{2} - x$ :

\frac{\cos x}{1 + \sin x} = \frac{\sin\phi}{1 + \cos\phi} = \frac{2\sin\tfrac{\phi}{2}\cos\tfrac{\phi}{2}}{2\cos^2\tfrac{\phi}{2}} = \tan\frac{\phi}{2} = \tan\!\left(\frac{\pi}{4} - \frac{x}{2}\right)

Here we used the half-angle identity $\dfrac{\sin\phi}{1 + \cos\phi} = \tan\dfrac{\phi}{2}$ . Same answer, arguably faster once you know that identity cold.

The identity $\dfrac{\sin\theta}{1 + \cos\theta} = \tan\dfrac{\theta}{2}$ is a 5-second shortcut worth remembering. It appears in integration problems too — particularly $\int \dfrac{dx}{1 + \cos x}$ type integrals in Class 12.

Common Mistake

A very common error is flipping the sign and writing the answer as $\dfrac{x}{2} - \dfrac{\pi}{4}$ instead of $\dfrac{\pi}{4} - \dfrac{x}{2}$ .

This happens when students divide by $\cos\tfrac{x}{2}$ but forget which term was on top. The numerator after cancellation is $\cos\tfrac{x}{2} - \sin\tfrac{x}{2}$ (cosine first, sine second), which gives $1 - \tan\tfrac{x}{2}$ in the numerator. That matches $\tan(\pi/4 - x/2)$ , not $\tan(x/2 - \pi/4)$ . The sign swap costs the full 3 marks in CBSE.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next