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CDS Mathematics · Trigonometric Ratios and Identities

Eliminating θ & Substitution Chains

When two equations define p and q through the same angle, find the relation between p and q that no longer mentions θ — by squaring and adding, by rewriting in sine and cosine, or by substituting a given relation into itself.

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Why this matters

Thirty-four PYQs and the hardest page in the chapter — sixteen are HARD, and the 2026 papers set four of these as linked pairs. Each looks unique, but they use only four moves. Recognise which one the question is built on and the algebra is short.

Concept 1 of 4: Square and add

If x=acos⁡θ+bsin⁡θx = a\cos\theta + b\sin\theta and y=asin⁡θ−bcos⁡θy = a\sin\theta - b\cos\theta, squaring both makes the cross terms appear with opposite signs, and adding cancels them. What survives is a combination of sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta, which is 11 — and θ\theta has gone.

Definition

Square each given relation and add (or subtract) so the cross terms cancel:

  • (acos⁡θ+bsin⁡θ)2+(asin⁡θ−bcos⁡θ)2=a2+b2(a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 = a^2 + b^2;
  • with secant and tangent, subtract instead, so sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1 does the work: (msec⁡A+ntan⁡A)2−(mtan⁡A+nsec⁡A)2=m2−n2(m\sec A + n\tan A)^2 - (m\tan A + n\sec A)^2 = m^2 - n^2;
  • relations like p+qcot⁡θ=3cosec⁡θp + q\cot\theta = 3\operatorname{cosec}\theta and q−pcot⁡θ=2cosec⁡θq - p\cot\theta = 2\operatorname{cosec}\theta square and add to (p2+q2)(1+cot⁡2θ)=13cosec⁡2θ(p^2 + q^2)(1 + \cot^2\theta) = 13\operatorname{cosec}^2\theta, so p2+q2=13p^2 + q^2 = 13.

The cancelling squares

(acos⁡θ+bsin⁡θ)2+(asin⁡θ−bcos⁡θ)2=a2+b2(a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 = a^2 + b^2

Worked example

If x=3sec⁡θ+2tan⁡θx = 3\sec\theta + 2\tan\theta and y=3tan⁡θ+2sec⁡θy = 3\tan\theta + 2\sec\theta, find x2−y2x^2 - y^2.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2017 · CDS (I) 2017 — Elementary Mathematics · Q68Easy

Example 1 · Trigonometric Ratios and Identities · Eliminating θ and Substitution Chains

If x=acos⁡θ+bsin⁡θx = a \cos \theta + b \sin \theta and y=asin⁡θ−bcos⁡θy = a \sin \theta - b \cos \theta, then what is x2+y2x^2 + y^2 equal to ?

Add for sine–cosine, subtract for secant–tangent

With sine and cosine the useful identity is a sum (sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1); with secant and tangent it is a difference (sec⁡2−tan⁡2=1\sec^2 - \tan^2 = 1). Choosing the wrong operation leaves the cross terms doubled instead of cancelled.
Drill 6 more on square and add

Concept 2 of 4: Rewrite each quantity in sine and cosine, then combine

cosec⁡θ−sin⁡θ\operatorname{cosec}\theta - \sin\theta looks awkward until it is written as one fraction: 1−sin⁡2θsin⁡θ=cos⁡2θsin⁡θ\dfrac{1 - \sin^2\theta}{\sin\theta} = \dfrac{\cos^2\theta}{\sin\theta}. Do that to each given quantity, and their product, quotient or sum is a clean expression in sin⁡θcos⁡θ\sin\theta\cos\theta — often a constant.

Definition

  • cosec⁡θ−sin⁡θ=cos⁡2θsin⁡θ\operatorname{cosec}\theta - \sin\theta = \dfrac{\cos^2\theta}{\sin\theta} and sec⁡θ−cos⁡θ=sin⁡2θcos⁡θ\sec\theta - \cos\theta = \dfrac{\sin^2\theta}{\cos\theta}.
  • Their product is sin⁡θcos⁡θ\sin\theta\cos\theta; their quotient is cot⁡3θ\cot^3\theta or tan⁡3θ\tan^3\theta.
  • cot⁡θ(1±sin⁡θ)=cot⁡θ±cos⁡θ\cot\theta(1 \pm \sin\theta) = \cot\theta \pm \cos\theta: their product is cot⁡2θcos⁡2θ\cot^2\theta\cos^2\theta, their sum 2cot⁡θ2\cot\theta, their difference 2cos⁡θ2\cos\theta.
  • If the given quantities are p3p^3 and q3q^3, expect answers in p2p^2, q2q^2 — that is, (sin⁡θcos⁡θ)2/3(\sin\theta\cos\theta)^{2/3}.

The two reductions that recur

cosec⁡θ−sin⁡θ=cos⁡2θsin⁡θ,sec⁡θ−cos⁡θ=sin⁡2θcos⁡θ\operatorname{cosec}\theta - \sin\theta = \frac{\cos^2\theta}{\sin\theta}, \qquad \sec\theta - \cos\theta = \frac{\sin^2\theta}{\cos\theta}

Worked example

If cosec⁡θ−sin⁡θ=a\operatorname{cosec}\theta - \sin\theta = a and sec⁡θ−cos⁡θ=b\sec\theta - \cos\theta = b, find a2b2(a2+b2+3)a^2b^2(a^2 + b^2 + 3).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (II) 2018 — Elementary Mathematics · Q70Moderate

Example 2 · Trigonometric Ratios and Identities · Eliminating θ and Substitution Chains

If cot⁡θ(1+sin⁡θ)=4m\cot \theta (1 + \sin \theta) = 4m and cot⁡θ(1−sin⁡θ)=4n\cot \theta (1 - \sin \theta) = 4n, then which one of the following is correct ?

Square roots need the sign of θ's quadrant

cot⁡2θcos⁡2θ\sqrt{\cot^2\theta\cos^2\theta} is cot⁡θcos⁡θ\cot\theta\cos\theta only when that product is positive. The stems that use it restrict θ\theta to 0<θ<90∘0 < \theta < 90^\circ for exactly this reason.
Drill 14 more on rewrite each quantity in sine and cosine, then combine

Concept 3 of 4: Substitution chains: sin x + sin²x = 1

sin⁡x+sin⁡2x=1\sin x + \sin^2 x = 1 says sin⁡x=1−sin⁡2x=cos⁡2x\sin x = 1 - \sin^2 x = \cos^2 x. That one swap is the whole question: every cos⁡2x\cos^2 x in the target can be replaced by sin⁡x\sin x, and the target turns back into the given relation. The powers — cos⁡12\cos^{12}, cos⁡10\cos^{10} — are there to hide a binomial expansion.

Definition

  • From sin⁡x+sin⁡2x=1\sin x + \sin^2 x = 1: sin⁡x=cos⁡2x\sin x = \cos^2 x. (From cos⁡x+cos⁡2x=1\cos x + \cos^2 x = 1: cos⁡x=sin⁡2x\cos x = \sin^2 x.)
  • Look for (u+1)n(u + 1)^n or uk(u+1)nu^k(u + 1)^n in the target: cos⁡6x+3cos⁡4x+3cos⁡2x+1\cos^6 x + 3\cos^4 x + 3\cos^2 x + 1 is (cos⁡2x+1)3(\cos^2 x + 1)^3, and cos⁡12x+3cos⁡10x+3cos⁡8x+cos⁡6x=cos⁡6x(cos⁡2x+1)3\cos^{12} x + 3\cos^{10} x + 3\cos^8 x + \cos^6 x = \cos^6 x(\cos^2 x + 1)^3.
  • Then cos⁡2x(cos⁡2x+1)=sin⁡x(sin⁡x+1)=1\cos^2 x(\cos^2 x + 1) = \sin x(\sin x + 1) = 1.
  • Nested squares: if tan⁡8θ+cot⁡8θ=m\tan^8\theta + \cot^8\theta = m, unwind with t2+1t2=(t+1t)2−2t^2 + \dfrac{1}{t^2} = \left(t + \dfrac1t\right)^2 - 2, one level at a time.

The swap and the grouping

sin⁡x+sin⁡2x=1⇒sin⁡x=cos⁡2x,cos⁡2x(cos⁡2x+1)=1\sin x + \sin^2 x = 1 \Rightarrow \sin x = \cos^2 x, \qquad \cos^2 x(\cos^2 x + 1) = 1

Worked example

If cos⁡x+cos⁡2x=1\cos x + \cos^2 x = 1, find sin⁡8x+2sin⁡6x+sin⁡4x\sin^8 x + 2\sin^6 x + \sin^4 x.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (II) 2018 — Elementary Mathematics · Q68Hard

Example 3 · Trigonometric Ratios and Identities · Eliminating θ and Substitution Chains

If sin⁡2x+sin⁡x=1\sin^2 x + \sin x = 1, then what is the value of cos⁡12x+3cos⁡10x+3cos⁡8x+cos⁡6x\cos^{12} x + 3 \cos^{10} x + 3 \cos^8 x + \cos^6 x ?

Swap the square, not the first power

sin⁡x+sin⁡2x=1\sin x + \sin^2 x = 1 gives sin⁡x=cos⁡2x\sin x = \cos^2 x — it does not give sin⁡2x=cos⁡x\sin^2 x = \cos x. That second form belongs to the cosine version of the question, and the two are set in alternate years.
Drill 5 more on substitution chains: sin x + sin²x = 1

Concept 4 of 4: Solving p sin²α + q cos²α = m for tan²α

psin⁡2α+qcos⁡2α=mp\sin^2\alpha + q\cos^2\alpha = m is a weighted average of pp and qq with weights sin⁡2α\sin^2\alpha and cos⁡2α\cos^2\alpha. Replace cos⁡2α\cos^2\alpha by 1−sin⁡2α1 - \sin^2\alpha, and sin⁡2α\sin^2\alpha comes out as a simple fraction; cos⁡2α\cos^2\alpha and tan⁡2α\tan^2\alpha follow.

Definition

From psin⁡2α+qcos⁡2α=mp\sin^2\alpha + q\cos^2\alpha = m:

  • sin⁡2α=m−qp−q\sin^2\alpha = \dfrac{m - q}{p - q} and cos⁡2α=p−mp−q\cos^2\alpha = \dfrac{p - m}{p - q};
  • so tan⁡2α=m−qp−m\tan^2\alpha = \dfrac{m - q}{p - m}.

Ratios of sines and cosines work the same way: if sin⁡α=ksin⁡β\sin\alpha = k\sin\beta and cos⁡α=lcos⁡β\cos\alpha = l\cos\beta, substitute both into sin⁡2α+cos⁡2α=1\sin^2\alpha + \cos^2\alpha = 1 to find sin⁡2β\sin^2\beta.

The weighted-average solution

psin⁡2α+qcos⁡2α=m  ⇒  tan⁡2α=m−qp−mp\sin^2\alpha + q\cos^2\alpha = m \;\Rightarrow\; \tan^2\alpha = \frac{m - q}{p - m}

Worked example

If 5sin⁡2α+2cos⁡2α=35\sin^2\alpha + 2\cos^2\alpha = 3, find tan⁡2α\tan^2\alpha.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2026 · CDS (I) 2026 — Elementary Mathematics · Q95Moderate

Example 4 · Trigonometric Ratios and Identities · Eliminating θ and Substitution Chains

for the items that follow : psin⁡2α+qcos⁡2α=mp\sin^2\alpha + q\cos^2\alpha = m and pcos⁡2β+qsin⁡2β=np\cos^2\beta + q\sin^2\beta = n
What is tan⁡2α\tan^2\alpha equal to ?

Keep the sign pattern of the fraction

tan⁡2α=m−qp−m\tan^2\alpha = \dfrac{m - q}{p - m}. The options include m−pq−m\dfrac{m - p}{q - m} and m−qm−p\dfrac{m - q}{m - p}, which differ by a sign in the denominator. A squared tangent cannot be negative, so check your answer's sign with sample values (p>m>qp > m > q).
Drill 5 more on solving p sin²α + q cos²α = m for tan²α

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 (4)

  • Square and add

    The cancelling squares

    (acos⁡θ+bsin⁡θ)2+(asin⁡θ−bcos⁡θ)2=a2+b2(a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 = a^2 + b^2
  • Rewrite each quantity in sine and cosine, then combine

    The two reductions that recur

    cosec⁡θ−sin⁡θ=cos⁡2θsin⁡θ,sec⁡θ−cos⁡θ=sin⁡2θcos⁡θ\operatorname{cosec}\theta - \sin\theta = \frac{\cos^2\theta}{\sin\theta}, \qquad \sec\theta - \cos\theta = \frac{\sin^2\theta}{\cos\theta}
  • Substitution chains: sin x + sin²x = 1

    The swap and the grouping

    sin⁡x+sin⁡2x=1⇒sin⁡x=cos⁡2x,cos⁡2x(cos⁡2x+1)=1\sin x + \sin^2 x = 1 \Rightarrow \sin x = \cos^2 x, \qquad \cos^2 x(\cos^2 x + 1) = 1
  • Solving p sin²α + q cos²α = m for tan²α

    The weighted-average solution

    psin⁡2α+qcos⁡2α=m  ⇒  tan⁡2α=m−qp−mp\sin^2\alpha + q\cos^2\alpha = m \;\Rightarrow\; \tan^2\alpha = \frac{m - q}{p - m}

Watch out for (4)

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