NDA Maths · Inverse Trigonometry

Solving Equations & Geometric Applications

Inverse-trig equations are solved by collapsing them with the complementary identity (or the sum formula) to a single inverse function, then checking the root is valid; geometric problems read angles as arctangents of height-over-distance.

All Inverse Trigonometry notes NDA Maths strategy

Why this matters

6 PYQs. Two reliable moves cover the subtopic: use sin⁻¹x + cos⁻¹x = π/2 to turn a mixed equation into one unknown, and watch the validity condition (ab < 1) when you apply the tan⁻¹ sum formula. Geometric questions are right-triangle arctangents.

Concept 1 of 2

Solving Inverse-Trig Equations

Intuition

Most equations mix two inverse functions. Replace one using the complementary identity so only a single unknown inverse remains, solve for it, then recover x. When the tan⁻¹ sum formula is used to form a quadratic, discard any root that violates its validity range.

Definition

Complementary collapse: in $a\sin^{-1}x + b\cos^{-1}x = c$ , write $\cos^{-1}x = \tfrac{\pi}{2} - \sin^{-1}x$ to get one unknown.
Sum-formula equations: $\tan^{-1}f(x) + \tan^{-1}g(x) = \tfrac{\pi}{4}$ becomes $\dfrac{f+g}{1-fg} = 1$ ; solve, then **reject roots where $fg \ge 1$ ** or where the principal range is exceeded.
Existence: $\sin^{-1}x - \cos^{-1}x = k$ has a solution only when the implied $\sin^{-1}x = \tfrac{\pi/2 + k}{2}$ lands in range.

Collapse with the complementary identity

a\sin^{-1}x + b\cos^{-1}x = c \ \xrightarrow{\cos^{-1}x = \frac{\pi}{2}-\sin^{-1}x}\ \text{one unknown}

Worked example

Solve

\sin^{-1}x + \cos^{-1}(1-x) = \tfrac{\pi}{2}

Compare with $\sin^{-1}u + \cos^{-1}u = \tfrac{\pi}{2}$ : this holds exactly when the two arguments are equal, $x = 1 - x$ .
$2x = 1$ .

Answer:

x = \tfrac{1}{2}

From the bank · past-year question

Example 1Inverse TrigonometryMODERATE

3\sin^{-1}x+\cos^{-1}x=\pi

, then what is

x

equal to?

[Q37 · Apr · 2022]

Reject roots that break the sum-formula validity

Forming

\frac{f+g}{1-fg}=1

can introduce a root where

fg>1

— there the real sum is

\tfrac{\pi}{4}+\pi

, not

\tfrac{\pi}{4}

. Always test each algebraic root against the original equation.

Drill 3 more on solving inverse-trig equations

Concept 2 of 2

Geometric Applications

Intuition

An angle of elevation is just the arctangent of height over horizontal distance. When a problem asks for the angle subtended between two heights, take the difference of two arctangents — exactly the sum/difference formula again.

Definition

Angle of elevation of a point at height $h$ , horizontal distance $d$ : $\theta = \tan^{-1}\dfrac{h}{d}$ .
Angle subtended by a segment between heights $h_1 < h_2$ at distance $d$ : $\tan^{-1}\dfrac{h_2}{d} - \tan^{-1}\dfrac{h_1}{d} = \tan^{-1}\dfrac{(h_2-h_1)d}{d^2 + h_1 h_2}$ .
Conditions linking $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ in AP with $x,y,z$ in GP typically force $x=y=z$ .

Subtended angle

\tan^{-1}\dfrac{h_2}{d} - \tan^{-1}\dfrac{h_1}{d} = \tan^{-1}\dfrac{(h_2-h_1)\,d}{d^2 + h_1 h_2}

Worked example

From a point 12 m from a pole, the top (height 9 m) and a flag at 21 m subtend what extra angle? Find

\tan(\angle)

between the 9 m and 21 m marks.

Angles: $\alpha = \tan^{-1}\tfrac{9}{12} = \tan^{-1}\tfrac34$ , $\beta = \tan^{-1}\tfrac{21}{12} = \tan^{-1}\tfrac74$ .
$\tan(\beta - \alpha) = \dfrac{\frac74 - \frac34}{1 + \frac74\cdot\frac34} = \dfrac{1}{1 + \frac{21}{16}} = \dfrac{16}{37}$ .

Answer:

\tan(\angle) = \dfrac{16}{37}

From the bank · past-year question

Example 2Inverse TrigonometryHARD

A man at M, standing 100 m away from the base (P) of a chimney of height 50 m, observes the angle of elevation of the highest point (Q) of the smoke to be 45°. The highest point of the chimney is at R. Further P, R and Q are in a straight line and the straight line is perpendicular to PM. What is the angle RMQ equal to?

[Q48 · Apr · 2025]

Drill 1 more on geometric applications

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Solving Inverse-Trig Equations
Collapse with the complementary identity
$a\sin^{-1}x + b\cos^{-1}x = c \ \xrightarrow{\cos^{-1}x = \frac{\pi}{2}-\sin^{-1}x}\ \text{one unknown}$
Geometric Applications
Subtended angle
$\tan^{-1}\dfrac{h_2}{d} - \tan^{-1}\dfrac{h_1}{d} = \tan^{-1}\dfrac{(h_2-h_1)\,d}{d^2 + h_1 h_2}$

Watch out for (1)

Reject roots that break the sum-formula validity
→ Solving Inverse-Trig Equations

Mastery check — 4 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Inverse TrigonometryMODERATE

\tan^{-1}\left(\frac{1}{2}\right)+\tan^{-1}\left(\frac{x}{3}\right)=\frac{\pi}{4}

, where

0<x<6

, then what is

x

equal to?

[Q36 · Apr · 2022]

Example 2Inverse TrigonometryHARD

Let

x

y

z

be positive real numbers such that

x

y

z

are in GP and

\tan^{-1}x

\tan^{-1}y

and

\tan^{-1}z

are in AP. Then which one of the following is correct?

[Q47 · Apr · 2017]

Example 3Inverse TrigonometryMODERATE

Consider the following values of x: 1. 8 2.

-4

\frac{1}{6}

-\frac{1}{4}

Which of the above values of x is/are the solution(s) of the equation

\tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}

[Q45 · Sep · 2018]

Example 4Inverse TrigonometryEASY

The equation

\sin^{-1} x - \cos^{-1} x = \frac{\pi}{6}

has

[Q21 · Apr · 2021]

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