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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.

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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.

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Concept 1 of 2: Solving Inverse-Trig Equations

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 asin⁡−1x+bcos⁡−1x=ca\sin^{-1}x + b\cos^{-1}x = c, write cos⁡−1x=π2−sin⁡−1x\cos^{-1}x = \tfrac{\pi}{2} - \sin^{-1}x to get one unknown.
  • Sum-formula equations: tan⁡−1f(x)+tan⁡−1g(x)=π4\tan^{-1}f(x) + \tan^{-1}g(x) = \tfrac{\pi}{4} becomes f+g1−fg=1\dfrac{f+g}{1-fg} = 1; solve, then reject roots where fg≥1fg \ge 1 or where the principal range is exceeded.
  • Existence: sin⁡−1x−cos⁡−1x=k\sin^{-1}x - \cos^{-1}x = k has a solution only when the implied sin⁡−1x=π/2+k2\sin^{-1}x = \tfrac{\pi/2 + k}{2} lands in range.

Collapse with the complementary identity

asin⁡−1x+bcos⁡−1x=c →cos⁡−1x=π2−sin⁡−1x one unknowna\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⁡−1x+cos⁡−1(1−x)=π2\sin^{-1}x + \cos^{-1}(1-x) = \tfrac{\pi}{2}.
Practice this concept2 quick reps

The same idea in a real exam question:

NDA · Apr 2022 · Q37Moderate

Example 1 · Inverse Trigonometry · Solving Inverse Trigonometric Equations and Geometric Applications

If 3sin⁡−1x+cos⁡−1x=π3\sin^{-1}x+\cos^{-1}x=\pi, then what is xx equal to?

Reject roots that break the sum-formula validity

Forming f+g1−fg=1\frac{f+g}{1-fg}=1 can introduce a root where fg>1fg>1 — there the real sum is π4+π\tfrac{\pi}{4}+\pi, not π4\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

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 hh, horizontal distance dd: θ=tan⁡−1hd\theta = \tan^{-1}\dfrac{h}{d}.
  • Angle subtended by a segment between heights h1<h2h_1 < h_2 at distance dd: tan⁡−1h2d−tan⁡−1h1d=tan⁡−1(h2−h1)dd2+h1h2\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⁡−1x,tan⁡−1y,tan⁡−1z\tan^{-1}x, \tan^{-1}y, \tan^{-1}z in AP with x,y,zx,y,z in GP typically force x=y=zx=y=z.

Subtended angle

tan⁡−1h2d−tan⁡−1h1d=tan⁡−1(h2−h1) dd2+h1h2\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⁡(∠)\tan(\angle) between the 9 m and 21 m marks.
Practice this concept2 quick reps

The same idea in a real exam question:

NDA · Apr 2025 · Q48Hard

Example 2 · Inverse Trigonometry · Solving Inverse Trigonometric Equations and Geometric Applications

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?

Subtended-angle denominator is d² + h₁h₂

For the angle between two heights at distance dd: tan⁡−1h2d−tan⁡−1h1d=tan⁡−1(h2−h1)dd2+h1h2\tan^{-1}\dfrac{h_2}{d}-\tan^{-1}\dfrac{h_1}{d}=\tan^{-1}\dfrac{(h_2-h_1)d}{d^2+h_1h_2}. The denominator is d2+h1h2d^2+h_1h_2 (the 1+ab1+ab of the difference formula), not d2−h1h2d^2-h_1h_2. Getting that sign wrong flips the answer.
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

    asin⁡−1x+bcos⁡−1x=c →cos⁡−1x=π2−sin⁡−1x one unknowna\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⁡−1h2d−tan⁡−1h1d=tan⁡−1(h2−h1) dd2+h1h2\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 (2)

Test yourself on Inverse Trigonometry

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