Class 12 Maths, Worked in Full Past papersGet updates
CBSE Sample Paper 2020-21 · Part B: Section III · 2 marks

Q19: Simplest form of tan⁡−1(cos⁡x1−sin⁡x)\tan^{-1}\left(\frac{\cos x}{1 - \sin x}\right)⁠, −3π2<x<π2-\frac{3\pi}{2} < x < \frac{\pi}{2}

Sample 2020-21Ch 2 · Inverse Trigonometric Functions
The question

Express tan⁡−1(cos⁡x1−sin⁡x)\tan^{-1}\left(\frac{\cos x}{1 - \sin x}\right)⁠, −3π2<x<π2\frac{-3\pi}{2} < x < \frac{\pi}{2} in the simplest form.

Exam answer

Let θ=π4−x2\theta = \frac{\pi}{4} - \frac{x}{2}⁠. Then π2−x=2θ\frac{\pi}{2} - x = 2\theta⁠, and

−3π2<x<π2⇒0<θ<π-\frac{3\pi}{2} < x < \frac{\pi}{2} \Rightarrow 0 < \theta < \pi
Why?

The plan is to turn the fraction into tan⁡\tan of an angle, so that tan⁡−1\tan^{-1} and tan⁡\tan undo each other. The half angle θ\theta is chosen so that double-angle formulas apply to π2−x\frac{\pi}{2} - x⁠.

The range of θ\theta comes from the given range of xx⁠: halve it, multiply by −1-1 (which reverses the signs), then add π4\frac{\pi}{4}⁠.

−3π4<x2<π43π4>−x2>−π4π>π4−x2>0\begin{aligned} -\frac{3\pi}{4} &< \frac{x}{2} < \frac{\pi}{4} \\ \frac{3\pi}{4} &> -\frac{x}{2} > -\frac{\pi}{4} \\ \pi &> \frac{\pi}{4} - \frac{x}{2} > 0 \end{aligned}

Now cos⁡x=sin⁡(π2−x)=sin⁡2θ\cos x = \sin\left(\frac{\pi}{2} - x\right) = \sin 2\theta and sin⁡x=cos⁡(π2−x)=cos⁡2θ\sin x = \cos\left(\frac{\pi}{2} - x\right) = \cos 2\theta⁠. So

tan⁡−1(cos⁡x1−sin⁡x)=tan⁡−1[sin⁡2θ1−cos⁡2θ]=tan⁡−1[2sin⁡θcos⁡θ2sin⁡2θ]=tan⁡−1(cot⁡θ)=tan⁡−1[tan⁡(π2−θ)]\begin{aligned} &\tan^{-1}\left(\frac{\cos x}{1 - \sin x}\right) \\ &= \tan^{-1}\left[\frac{\sin 2\theta}{1 - \cos 2\theta}\right] \\ &= \tan^{-1}\left[\frac{2\sin\theta\cos\theta}{2\sin^2\theta}\right] \\ &= \tan^{-1}(\cot\theta) \\ &= \tan^{-1}\left[\tan\left(\frac{\pi}{2} - \theta\right)\right] \end{aligned}
Why?

First, cos⁡x=sin⁡(π2−x)\cos x = \sin\left(\frac{\pi}{2} - x\right) and sin⁡x=cos⁡(π2−x)\sin x = \cos\left(\frac{\pi}{2} - x\right)⁠, because sine and cosine swap for complementary angles. Then, with π2−x=2θ\frac{\pi}{2} - x = 2\theta⁠, the double-angle formulas give

sin⁡2θ=2sin⁡θcos⁡θ1−cos⁡2θ=2sin⁡2θ\begin{aligned} \sin 2\theta &= 2\sin\theta\cos\theta \\ 1 - \cos 2\theta &= 2\sin^2\theta \end{aligned}

The second one comes from cos⁡2θ=1−2sin⁡2θ\cos 2\theta = 1 - 2\sin^2\theta⁠:

1−cos⁡2θ=1−(1−2sin⁡2θ)=1−1+2sin⁡2θ=2sin⁡2θ\begin{aligned} 1 - \cos 2\theta &= 1 - (1 - 2\sin^2\theta) \\ &= 1 - 1 + 2\sin^2\theta \\ &= 2\sin^2\theta \end{aligned}

Since 0<θ<π0 < \theta < \pi⁠, sin⁡θ≠0\sin\theta \ne 0⁠, so 2sin⁡θ2\sin\theta cancels and leaves cos⁡θsin⁡θ=cot⁡θ\frac{\cos\theta}{\sin\theta} = \cot\theta⁠. Finally cot⁡θ=tan⁡(π2−θ)\cot\theta = \tan\left(\frac{\pi}{2} - \theta\right)⁠.

2sin⁡θcos⁡θ2sin⁡2θ=2sin⁡θ⋅cos⁡θ2sin⁡θ⋅sin⁡θ=cos⁡θsin⁡θ=cot⁡θ\begin{aligned} \frac{2\sin\theta\cos\theta}{2\sin^2\theta} &= \frac{2\sin\theta \cdot \cos\theta}{2\sin\theta \cdot \sin\theta} \\ &= \frac{\cos\theta}{\sin\theta} = \cot\theta \end{aligned}

Since 0<θ<π0 < \theta < \pi⁠, −π2<π2−θ<π2-\frac{\pi}{2} < \frac{\pi}{2} - \theta < \frac{\pi}{2}⁠. So

Why?

tan⁡−1\tan^{-1} gives values only in its principal range, so

tan⁡−1(tan⁡u)=u\tan^{-1}(\tan u) = u

holds only when −π2<u<π2-\frac{\pi}{2} < u < \frac{\pi}{2}⁠. That is why the range of θ\theta was needed. From 0<θ<π0 < \theta < \pi⁠: −π<−θ<0-\pi < -\theta < 0⁠, then add π2\frac{\pi}{2}⁠.

tan⁡−1[tan⁡(π2−θ)]=π2−θ=π2−π4+x2\begin{aligned} &\tan^{-1}\left[\tan\left(\frac{\pi}{2} - \theta\right)\right] \\ &= \frac{\pi}{2} - \theta = \frac{\pi}{2} - \frac{\pi}{4} + \frac{x}{2} \end{aligned}
Why?

Watch the sign: −θ=−(π4−x2)=−π4+x2-\theta = -\left(\frac{\pi}{4} - \frac{x}{2}\right) = -\frac{\pi}{4} + \frac{x}{2}⁠. Then π2−π4=π4\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}⁠.

π2−θ=π2−(π4−x2)=π2−π4+x2=2π4−π4+x2=π4+x2\begin{aligned} \frac{\pi}{2} - \theta &= \frac{\pi}{2} - \left(\frac{\pi}{4} - \frac{x}{2}\right) \\ &= \frac{\pi}{2} - \frac{\pi}{4} + \frac{x}{2} \\ &= \frac{2\pi}{4} - \frac{\pi}{4} + \frac{x}{2} \\ &= \frac{\pi}{4} + \frac{x}{2} \end{aligned}

Check at x=0x = 0⁠: tan⁡−1(11−0)=tan⁡−11=π4\tan^{-1}\left(\frac{1}{1 - 0}\right) = \tan^{-1}1 = \frac{\pi}{4}⁠, and π4+0=π4\frac{\pi}{4} + 0 = \frac{\pi}{4}⁠.

∴\therefore the simplest form is π4+x2\dfrac{\pi}{4} + \dfrac{x}{2}⁠.

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Simplest form of an inverse trigonometric function.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Inverse Trigonometric Functions in past papers: question 18 of 18 · see them all