(a) Simplify: $\tan^{-1}\left[\dfrac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right]$ , $0 < x < \pi/4$ .

OR

(b) Evaluate: $\tan\left[\sin^{-1} 1 - \cos^{-1}(-1/2)\right]$ .

Q: (a) Simplify: $\tan^{-1}\left[\dfrac{\cos 2x - \sin 2x}{\cos 2x + \sin 2x}\right]$, $0 < x < \pi/4$. OR (b) Evaluate: $\tan\left[\sin^{-1} 1 - \cos^{-1}(-1/2)\right]$.

Both alternatives reduce to a single standard value using inverse-trigonometric identities. **Option (a):** $\pi/4 - 2x$ - Divide numerator and denominator by $\cos 2x$: $(1 - \tan 2x)/(1 + \tan 2x) = \tan(\pi/4 - 2x)$. - Since $0 < x < \pi/4$, the angle π/4 − 2x lies in $(-\pi/4, \pi/4)$, inside the principal branch of tan⁻¹, so $\tan^{-1}[\tan(\pi/4 - 2x)]$ = **$\pi/4 - 2x$**. **Option (b):** $-1/\sqrt{3}$ - $\sin^{-1} 1 = \pi/2$ and $\cos^{-1}(-1/2) = 2\pi/3$ (the negative argument places the angle in the second quadrant). - The bracket becomes $\pi/2 - 2\pi/3 = -\pi/6$, so $\tan(-\pi/6)$ = **$-1/\sqrt{3}$**.

Q: Why must the domain $0 < x < \pi/4$ be given in part (a)?

It restricts $\pi/4 - 2x$ to $(-\pi/4, \pi/4)$, inside tan⁻¹'s principal branch $(-\pi/2, \pi/2)$, so the identity $\tan^{-1}[\tan \theta] = \theta$ applies directly without adjustment.

Q: What are the standard values of $\sin^{-1} 1$ and $\cos^{-1}(-1/2)$?

$\sin^{-1} 1 = \pi/2$ and $\cos^{-1}(-1/2) = 2\pi/3$. Because the argument of cos⁻¹ is negative, its value lies in the second quadrant, between π/2 and π.

Inverse Trigonometric Functions

Inverse Trigonometric Functions — Simplification and Evaluation

Official Answer

Both alternatives reduce to a single standard value using inverse-trigonometric identities.

Option (a): $\pi/4 - 2x$

Divide numerator and denominator by $\cos 2x$ : $(1 - \tan 2x)/(1 + \tan 2x) = \tan(\pi/4 - 2x)$ .
Since $0 < x < \pi/4$ , the angle π/4 − 2x lies in $(-\pi/4, \pi/4)$ , inside the principal branch of tan⁻¹, so $\tan^{-1}[\tan(\pi/4 - 2x)]$ = $\pi/4 - 2x$ .

Option (b): $-1/\sqrt{3}$

$\sin^{-1} 1 = \pi/2$ and $\cos^{-1}(-1/2) = 2\pi/3$ (the negative argument places the angle in the second quadrant).
The bracket becomes $\pi/2 - 2\pi/3 = -\pi/6$ , so $\tan(-\pi/6)$ = $-1/\sqrt{3}$ .

inverse trigonometric functionstan(A-B) identityprincipal value branchsin inversecos inversesimplification

Marking Scheme

11 mark: correctly reducing (a) to $\tan(\pi/4 - 2x)$ form, or identifying $\sin^{-1} 1 = \pi/2$ and $\cos^{-1}(-1/2) = 2\pi/3$ in (b).
21 mark: correct final value — $\pi/4 - 2x$ (a) with branch justification, or $-1/\sqrt{3}$ (b).

Hint

Divide numerator and denominator of (a) by $\cos 2x$ to get $\tan(\pi/4 - 2x)$ ; for (b) recall $\sin^{-1} 1 = \pi/2$ and $\cos^{-1}(-1/2) = 2\pi/3$ .

Quick Oral Answer

Part (a) simplifies to $\pi/4 - 2x$ via $\tan(A-B)$ ; part (b) evaluates to $-1/\sqrt{3}$ using $\sin^{-1} 1 = \pi/2$ and $\cos^{-1}(-1/2) = 2\pi/3$ , giving $\tan(-\pi/6)$ .

Analysis & Explanation

This question tests fluency with standard inverse-trigonometric identities and principal-value ranges.

Concept: Part (a) uses the $\tan(A - B)$ expansion in reverse — dividing by cos 2x converts the ratio into $1$ and $\tan 2x$ . Part (b) requires the standard values $\sin^{-1} 1 = \pi/2$ and $\cos^{-1}(-1/2) = 2\pi/3$ .

Key point on (b): The inverse cosine of a negative number lies in the second quadrant $(\pi/2, \pi]$ , so $\cos^{-1}(-1/2) = 2\pi/3$ , not $\pi/3$ .

Exam trap: In part (a) students forget to confirm the simplified angle lies within tan⁻¹'s principal branch $(-\pi/2, \pi/2)$ ; the given domain $0 < x < \pi/4$ exists precisely to guarantee this.

Real-world link: Such simplifications resemble phase-difference calculations in wave optics and signal processing, where phase angles are expressed as ratios of trigonometric terms.

Common Mistakes

1Ignoring the given domain $0 < x < \pi/4$ and giving an angle outside the principal branch of tan⁻¹ in part (a).
2Taking $\cos^{-1}(-1/2)$ as $\pi/3$ instead of $2\pi/3$ — inverse cosine of a negative value lies in the second quadrant $(\pi/2, \pi]$ .
3Sign slip in $\tan(-\pi/6)$ , giving $+1/\sqrt{3}$ instead of the correct $-1/\sqrt{3}$ .

Interesting Facts

Principal value branches of inverse trigonometric functions were standardised specifically to make functions like tan⁻¹ single-valued and continuous.

The $\tan(A-B)$ identity used here is the same one applied in navigation and surveying to compute bearings and angles of elevation.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

Why must the domain $0 < x < \pi/4$ be given in part (a)?

It restricts $\pi/4 - 2x$ to $(-\pi/4, \pi/4)$ , inside tan⁻¹'s principal branch $(-\pi/2, \pi/2)$ , so the identity $\tan^{-1}[\tan \theta] = \theta$ applies directly without adjustment.

What are the standard values of $\sin^{-1} 1$ and $\cos^{-1}(-1/2)$ ?

$\sin^{-1} 1 = \pi/2$ and $\cos^{-1}(-1/2) = 2\pi/3$ . Because the argument of cos⁻¹ is negative, its value lies in the second quadrant, between π/2 and π.