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NDA Maths · Trigonometric Equations

Specific Forms — Vieta, Products & Logarithms

A recurring set of disguises: trig ratios appearing as the roots of a quadratic (use Vieta's relations), product equations that collapse via tan(A+B), and logarithmic trig equations solved by a substitution.

All Trigonometric Equations notesNDA Maths strategy

Why this matters

13 PYQs. These look different but each has a signature move — Vieta's sum/product when trig values are roots, the tan-sum identity for (1+tan)(1+tan) products, and t + 1/t for logarithmic pairs. Recognise the form and the solution is short.

Test yourself — a quick 14-question Trigonometric Equations recall quiz, instant score.

Concept 1 of 3: Trig Values as Roots of a Quadratic (Vieta)

When sin θ and cos θ (or tan α and tan β) are the roots of a quadratic, Vieta's relations hand you their sum and product directly — and a trig identity (like sin²+cos²=1) connects those to the coefficients without ever finding the angles.

Definition

If the trig values are roots of ax2+bx+c=0ax^2 + bx + c = 0, then sum =−ba= -\tfrac{b}{a} and product =ca= \tfrac{c}{a}. Combine with an identity:

  • sin⁡θ,cos⁡θ\sin\theta, \cos\theta roots: (sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ(\sin\theta+\cos\theta)^2 = 1 + 2\sin\theta\cos\theta gives a relation among a,b,ca,b,c (here a2−b2+2ac=0a^2 - b^2 + 2ac = 0).
  • tan⁡α,tan⁡β\tan\alpha, \tan\beta roots: tan⁡(α+β)=sum1−product\tan(\alpha+\beta) = \dfrac{\text{sum}}{1 - \text{product}}.
  • cot⁡α,cot⁡β\cot\alpha, \cot\beta roots: cot⁡(α+β)=product−1sum\cot(\alpha+\beta) = \dfrac{\text{product} - 1}{\text{sum}}.

Vieta + tan-sum

tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β=−b/a1−c/a\tan(\alpha+\beta) = \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} = \dfrac{-b/a}{1-c/a}

Worked example

If tan⁡α,tan⁡β\tan\alpha, \tan\beta are roots of x2−5x+6=0x^2 - 5x + 6 = 0, find tan⁡(α+β)\tan(\alpha+\beta).

The same idea in a real exam question:

NDA · Apr 2019 · Q46Moderate

Example 1 · Trigonometric Equations · Solving Specific Forms — Double-Angle, Product, Logarithmic, and Vieta

If the roots of the equation x2+px+q=0x^2 + px + q = 0 are tan⁡19°\tan19° and tan⁡26°\tan26°, then which one of the following is correct?

Vieta's product is c/ac/a, not −c/a-c/a

For ax2+bx+c=0ax^2+bx+c=0 the sum of roots is −ba-\tfrac{b}{a} (sign flips) but the product is +ca+\tfrac{c}{a} (no sign flip). Putting a minus on the product, or forgetting it on the sum, breaks the tan⁡(α+β)\tan(\alpha+\beta) computation.
Drill 3 more on trig values as roots of a quadratic (vieta)

Concept 2 of 3: Product & Sum-to-Product Forms

A product like (1+tan θ)(1+tan 9θ)=2 expands into exactly the numerator/denominator of the tan-sum formula — so it secretly says tan(10θ)=1. Sum-to-product turns sums of sines and cosines into products you can divide.

Definition

  • (1+tan A)(1+tan B) = 2 expands to tan⁡A+tan⁡B=1−tan⁡Atan⁡B\tan A + \tan B = 1 - \tan A\tan B, i.e. tan⁡(A+B)=1\tan(A+B) = 1, so A+B=π4A + B = \tfrac{\pi}{4} (this is the classic 45∘45^\circ identity).
  • Sum-to-product: sin⁡x+sin⁡y=2sin⁡x+y2cos⁡x−y2\sin x + \sin y = 2\sin\tfrac{x+y}{2}\cos\tfrac{x-y}{2}, cos⁡y−cos⁡x=2sin⁡x+y2sin⁡x−y2\cos y - \cos x = 2\sin\tfrac{x+y}{2}\sin\tfrac{x-y}{2}; dividing isolates tan⁡x−y2\tan\tfrac{x-y}{2}.
  • tan(45° + θ) = 1 + sin 2θ-type equations: expand both sides in tan⁡θ\tan\theta and solve the resulting algebraic equation.

The product identity

(1+tan⁡A)(1+tan⁡B)=2  ⟺  A+B=π4(1+\tan A)(1+\tan B) = 2 \iff A + B = \tfrac{\pi}{4}

Worked example

If (1+tan⁡θ)(1+tan⁡(45∘−θ))=k(1+\tan\theta)(1+\tan(45^\circ - \theta)) = k, find kk.

The same idea in a real exam question:

NDA · Apr 2022 · Q39Moderate

Example 2 · Trigonometric Equations · Solving Specific Forms — Double-Angle, Product, Logarithmic, and Vieta

If (1+tan⁡θ)(1+tan⁡9θ)=2(1+\tan\theta)(1+\tan9\theta)=2, then what is the value of tan⁡(10θ)\tan(10\theta)?
Drill 4 more on product & sum-to-product forms

Concept 3 of 3: Logarithmic & Special Trig Equations

A logarithm with a trig base, log_{cos x} sin x, is just an exponent equation in disguise. When two reciprocal logs add to 2, the t + 1/t = 2 trick forces t = 1, collapsing it to cos x = sin x.

Definition

  • log⁡cos⁡xsin⁡x=1\log_{\cos x}\sin x = 1 means sin⁡x=cos⁡x\sin x = \cos x, so x=π4x = \tfrac{\pi}{4} (in the first quadrant).
  • log⁡sin⁡xcos⁡x+log⁡cos⁡xsin⁡x=2\log_{\sin x}\cos x + \log_{\cos x}\sin x = 2: the two terms are reciprocals t+1tt + \tfrac1t, and t+1t=2⇒t=1t + \tfrac1t = 2 \Rightarrow t = 1, giving sin⁡x=cos⁡x\sin x = \cos x.
  • Special-angle outputs: equations reducing to sin⁡2θ=cos⁡3θ\sin 2\theta = \cos 3\theta give θ=18∘\theta = 18^\circ, where sin⁡18∘=5−14\sin 18^\circ = \tfrac{\sqrt5 - 1}{4} — a value worth memorising.

Reciprocal-log trick

t+1t=2  ⟺  t=1t + \tfrac{1}{t} = 2 \iff t = 1

Worked example

Solve log⁡cos⁡xsin⁡x=1\log_{\cos x}\sin x = 1 for 0<x<π20 < x < \tfrac{\pi}{2}.

The same idea in a real exam question:

NDA · Apr 2026 · Q38Moderate

Example 3 · Trigonometric Equations · Solving Specific Forms — Double-Angle, Product, Logarithmic, and Vieta

What is the smallest positive xx satisfying log⁡sin⁡xcos⁡x+log⁡cos⁡xsin⁡x=2\log_{\sin x}\cos x+\log_{\cos x}\sin x=2?

A log base must be positive and ≠1\ne 1

In log⁡cos⁡xsin⁡x\log_{\cos x}\sin x the base cos⁡x\cos x must satisfy cos⁡x>0\cos x > 0 and cos⁡x≠1\cos x \ne 1, and the argument needs sin⁡x>0\sin x > 0. After solving tan⁡x=1\tan x = 1, keep only roots in the first quadrant — x=π4x = \tfrac{\pi}{4} — and reject any where the base/argument condition fails.
Drill 3 more on logarithmic & special trig equations

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 (3)

  • Trig Values as Roots of a Quadratic (Vieta)

    Vieta + tan-sum

    tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β=−b/a1−c/a\tan(\alpha+\beta) = \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} = \dfrac{-b/a}{1-c/a}
  • Product & Sum-to-Product Forms

    The product identity

    (1+tan⁡A)(1+tan⁡B)=2  ⟺  A+B=π4(1+\tan A)(1+\tan B) = 2 \iff A + B = \tfrac{\pi}{4}
  • Logarithmic & Special Trig Equations

    Reciprocal-log trick

    t+1t=2  ⟺  t=1t + \tfrac{1}{t} = 2 \iff t = 1

Watch out for (2)

Test yourself on Trigonometric Equations

10 past NDA questions from this chapter, timed at 13 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Trigonometric Equations chapter testDrill all 13 questions