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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.

Concept 1 of 3

Trig Values as Roots of a Quadratic (Vieta)

Intuition

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).

From the bank · past-year question

Example 1Trigonometric EquationsMODERATE
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?

[Q46 · Apr · 2019]

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

Intuition

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.

From the bank · past-year question

Example 2Trigonometric EquationsMODERATE
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)?

[Q39 · Apr · 2022]

Drill 4 more on product & sum-to-product forms

Concept 3 of 3

Logarithmic & Special Trig Equations

Intuition

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}.

From the bank · past-year question

Example 3Trigonometric EquationsMODERATE
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?

[Q38 · Apr · 2026]

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)

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