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SSC CGL Maths · Trigonometric Ratios and Identities

Reciprocal Pairs and Given Sums

Given one combination such as sec θ + tan θ or sin θ + cos θ, an identity hands you its partner, and the partner gives every other value asked for.

All Trigonometric Ratios and Identities notes

Why this matters

45 questions in the bank, 17 of them MODERATE. Eighteen turn on the sec θ ± tan θ or cosec θ ± cot θ pair, which appears in every year of the bank; most of the rest give a sum and want a square or a higher power. All of them reward one recognition, and none needs the angle itself.

Concept 1 of 4: sec θ + tan θ and sec θ − tan θ multiply to 1

sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1 is a difference of two squares, so it factors as (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1. Know one bracket and the other is its reciprocal. cosec and cot behave the same way.

Definition

  • If sec⁡θ+tan⁡θ=p\sec\theta + \tan\theta = p, then sec⁡θ−tan⁡θ=1p\sec\theta - \tan\theta = \frac{1}{p}.
  • Add the two: sec⁡θ=12(p+1p)\sec\theta = \frac{1}{2}\left(p + \frac{1}{p}\right). Subtract: tan⁡θ=12(p−1p)\tan\theta = \frac{1}{2}\left(p - \frac{1}{p}\right). Divide: sin⁡θ=p2−1p2+1\sin\theta = \frac{p^2 - 1}{p^2 + 1}.
  • If cosec θ+cot⁡θ=m\text{cosec}\,\theta + \cot\theta = m, then cosec θ−cot⁡θ=1m\text{cosec}\,\theta - \cot\theta = \frac{1}{m} and cos⁡θ=m2−1m2+1\cos\theta = \frac{m^2 - 1}{m^2 + 1}.
  • The same pair hides in other forms: 1+sin⁡θcos⁡θ=sec⁡θ+tan⁡θ\frac{1 + \sin\theta}{\cos\theta} = \sec\theta + \tan\theta, 1+sin⁡θ1−sin⁡θ=sec⁡θ+tan⁡θ\sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} = \sec\theta + \tan\theta, 1+cos⁡θ1−cos⁡θ=cosec θ+cot⁡θ\sqrt{\frac{1 + \cos\theta}{1 - \cos\theta}} = \text{cosec}\,\theta + \cot\theta, and sin⁡θ1+cos⁡θ=cosec θ−cot⁡θ\frac{\sin\theta}{1 + \cos\theta} = \text{cosec}\,\theta - \cot\theta.
  • Fast route: p2+1p^2 + 1, p2−1p^2 - 1 and 2p2p are the sides of a right triangle. For p=5p = 5 that is 26, 24, 10, which is the 5, 12, 13 triangle, so sec⁡θ=135\sec\theta = \frac{13}{5} and tan⁡θ=125\tan\theta = \frac{12}{5} at sight.
  • Limit: the square-root forms assume the denominator ratio is positive (cos⁡θ>0\cos\theta > 0 for the sec pair), which holds for every acute angle.

The reciprocal pairs

(sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1(cosec θ+cot⁡θ)(cosec θ−cot⁡θ)=1sec⁡θ+tan⁡θ=p⇒sin⁡θ=p2−1p2+11+sin⁡θ1−sin⁡θ=sec⁡θ+tan⁡θ(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1 \qquad (\text{cosec}\,\theta + \cot\theta)(\text{cosec}\,\theta - \cot\theta) = 1 \qquad \sec\theta + \tan\theta = p \Rightarrow \sin\theta = \frac{p^2 - 1}{p^2 + 1} \qquad \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} = \sec\theta + \tan\theta

Worked example

If sec⁡θ+tan⁡θ=7\sec\theta + \tan\theta = 7, find sec⁡θ\sec\theta, tan⁡θ\tan\theta and cos⁡θ\cos\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 26 Sep 2025, Shift 1 · Q73Moderate

Example 1 · Trigonometric Ratios and Identities · Reciprocal Pairs and Given Sums

If tan⁡A+sec⁡A=3\tan A + \sec A = 3, find tan⁡A\tan A.

Taking the partner as p − 1 or −p

If sec⁡θ+tan⁡θ=p\sec\theta + \tan\theta = p, then sec⁡θ−tan⁡θ\sec\theta - \tan\theta is 1p\frac{1}{p}, because the two multiply to 1. It is not p−1p - 1, −p-p or p2−1p^2 - 1.

Swapping the sum and the difference

Adding the pair gives 2sec⁡θ2\sec\theta; subtracting gives 2tan⁡θ2\tan\theta. Halving p+1pp + \frac{1}{p} and calling it tan is the most common slip, and the result is always a printed option.
Drill 17 more on sec θ + tan θ and sec θ − tan θ multiply to 1

Concept 2 of 4: Squaring a given sum

(sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta, so a given sum hands you the product sin⁡θcos⁡θ\sin\theta\cos\theta, and that product unlocks the difference, the squares and tan θ + cot θ.

Definition

  • (sin⁡θ±cos⁡θ)2=1±2sin⁡θcos⁡θ(\sin\theta \pm \cos\theta)^2 = 1 \pm 2\sin\theta\cos\theta, and the two squares add to 2: (sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2=2(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2.
  • tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ\tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta}.
  • A number plus its reciprocal: if x+1x=kx + \frac{1}{x} = k, then x2+1x2=k2−2x^2 + \frac{1}{x^2} = k^2 - 2. Use it for cos⁡θ+sec⁡θ\cos\theta + \sec\theta and tan⁡θ+cot⁡θ\tan\theta + \cot\theta.
  • For a positive x, x+1x=2x + \frac{1}{x} = 2 only when x=1x = 1. So tan⁡θ+cot⁡θ=2\tan\theta + \cot\theta = 2 means θ = 45°, and sin⁡θ+cosec θ=2\sin\theta + \text{cosec}\,\theta = 2 means sin⁡θ=1\sin\theta = 1.
  • Fast route: when the given value is the extreme one (sin⁡θ+cos⁡θ=2\sin\theta + \cos\theta = \sqrt{2}, tan⁡θ+cot⁡θ=2\tan\theta + \cot\theta = 2), the angle is 45°. Put it in and read the answer.
  • Limit: sin⁡θ+cos⁡θ\sin\theta + \cos\theta can never exceed 2≈1.414\sqrt{2} \approx 1.414, and for a positive x, x+1xx + \frac{1}{x} is never below 2. A given value beyond these has no angle.

Squared sums

(sin⁡θ±cos⁡θ)2=1±2sin⁡θcos⁡θ(sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2=2x+1x=k⇒x2+1x2=k2−2tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ(\sin\theta \pm \cos\theta)^2 = 1 \pm 2\sin\theta\cos\theta \qquad (\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2 \qquad x + \frac{1}{x} = k \Rightarrow x^2 + \frac{1}{x^2} = k^2 - 2 \qquad \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta}

Worked example

If sin⁡θ−cos⁡θ=13\sin\theta - \cos\theta = \frac{1}{3} and θ is acute, find sin⁡θcos⁡θ\sin\theta\cos\theta and sin⁡θ+cos⁡θ\sin\theta + \cos\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 19 Sep 2025, Shift 3 · Q65Easy

Example 2 · Trigonometric Ratios and Identities · Reciprocal Pairs and Given Sums

If sin⁡x+cos⁡x=2\sin x + \cos x = \sqrt{2}, what is the value of sin⁡x−cos⁡x\sin x - \cos x?

Forgetting the 2 in the cross term

(sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta, not 1+sin⁡θcos⁡θ1 + \sin\theta\cos\theta. Dropping the 2 doubles the product you find, and the doubled value is usually printed.

Answering a given that cannot happen

sin⁡θ+cos⁡θ=3\sin\theta + \cos\theta = \sqrt{3} would need sin⁡θcos⁡θ=1\sin\theta\cos\theta = 1, but sin⁡θcos⁡θ\sin\theta\cos\theta is never more than 12\frac{1}{2}. Any sum above 2\sqrt{2} describes no real angle.
Drill 15 more on squaring a given sum

Concept 3 of 4: sin⁴θ + cos⁴θ and sin⁶θ + cos⁶θ

Treat sin⁡2θ\sin^2\theta and cos⁡2θ\cos^2\theta as two numbers that add to 1. The fourth and sixth powers then come from ordinary algebra on a sum of 1, and each answer depends only on the product sin⁡2θcos⁡2θ\sin^2\theta\cos^2\theta.

Definition

Write p=sin⁡2θcos⁡2θp = \sin^2\theta\cos^2\theta.

  • sin⁡4θ+cos⁡4θ=(sin⁡2θ+cos⁡2θ)2−2p=1−2p\sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 - 2p = 1 - 2p.
  • sin⁡6θ+cos⁡6θ=(sin⁡2θ+cos⁡2θ)3−3p(sin⁡2θ+cos⁡2θ)=1−3p\sin^6\theta + \cos^6\theta = (\sin^2\theta + \cos^2\theta)^3 - 3p(\sin^2\theta + \cos^2\theta) = 1 - 3p.
  • sin⁡4θ−cos⁡4θ=(sin⁡2θ−cos⁡2θ)(sin⁡2θ+cos⁡2θ)=sin⁡2θ−cos⁡2θ\sin^4\theta - \cos^4\theta = (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta) = \sin^2\theta - \cos^2\theta.
  • p lies from 0 to 14\frac{1}{4}, so sin⁡4θ+cos⁡4θ\sin^4\theta + \cos^4\theta lies from 12\frac{1}{2} to 1.
  • Fast route: if the options are plain numbers, the expression is the same at every angle. Put θ = 0° (sin 0, cos 1) and read the answer.
  • Limit: this works only when the answer is a constant. If the options contain θ, use the algebra or test two angles.

Higher powers of sin and cos

sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θsin⁡6θ+cos⁡6θ=1−3sin⁡2θcos⁡2θsin⁡4θ−cos⁡4θ=sin⁡2θ−cos⁡2θ\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta \qquad \sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta \qquad \sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta

Worked example

If sin⁡θcos⁡θ=13\sin\theta\cos\theta = \frac{1}{3}, find sin⁡6θ+cos⁡6θ\sin^6\theta + \cos^6\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 21 Sep 2025, Shift 1 · Q68Easy

Example 3 · Trigonometric Ratios and Identities · Reciprocal Pairs and Given Sums

If sin⁡A+cos⁡A=1\sin A + \cos A = 1, then find the value of sin⁡4A+cos⁡4A\sin^4 A + \cos^4 A.

Writing sin⁴θ + cos⁴θ = 1

Only the squares add to 1. The fourth powers add to 1−2sin⁡2θcos⁡2θ1 - 2\sin^2\theta\cos^2\theta, which is 12\frac{1}{2} at 45° and 1 only where sin or cos is 0.

Using θ = 0° when the options contain θ

Putting in one angle finds a constant answer. When the options are expressions in θ, one angle can make two of them equal, so test a second angle before choosing.
Drill 5 more on sin⁴θ + cos⁴θ and sin⁶θ + cos⁶θ

Concept 4 of 4: Substitution chains such as cos θ + cos²θ = 1

The condition cos⁡θ+cos⁡2θ=1\cos\theta + \cos^2\theta = 1 says cos⁡θ=1−cos⁡2θ=sin⁡2θ\cos\theta = 1 - \cos^2\theta = \sin^2\theta. From then on every cos⁡θ\cos\theta can be swapped for sin⁡2θ\sin^2\theta, and back, until the asked expression turns into the given one.

Definition

  • cos⁡θ+cos⁡2θ=1\cos\theta + \cos^2\theta = 1 gives cos⁡θ=sin⁡2θ\cos\theta = \sin^2\theta, so cos⁡2θ=sin⁡4θ\cos^2\theta = \sin^4\theta.
  • sin⁡θ+sin⁡2θ=1\sin\theta + \sin^2\theta = 1 gives sin⁡θ=cos⁡2θ\sin\theta = \cos^2\theta, so sin⁡2θ=cos⁡4θ\sin^2\theta = \cos^4\theta.
  • tan⁡4θ+tan⁡2θ=1\tan^4\theta + \tan^2\theta = 1 gives tan⁡2θsec⁡2θ=1\tan^2\theta\sec^2\theta = 1, that is sin⁡2θ=cos⁡4θ\sin^2\theta = \cos^4\theta: the same chain in disguise.
  • Method: rewrite the condition as one swap, apply the swap to the asked expression, and look for the given expression to appear again.
  • The angle here is not a table angle (cos θ is about 0.618), so do not hunt for it. The swap is the whole method.

The two chains

cos⁡θ+cos⁡2θ=1⇒cos⁡θ=sin⁡2θsin⁡θ+sin⁡2θ=1⇒sin⁡θ=cos⁡2θ\cos\theta + \cos^2\theta = 1 \Rightarrow \cos\theta = \sin^2\theta \qquad \sin\theta + \sin^2\theta = 1 \Rightarrow \sin\theta = \cos^2\theta

Worked example

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

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 23 Sep 2024, Shift 2 · Q68Easy

Example 4 · Trigonometric Ratios and Identities · Reciprocal Pairs and Given Sums

If cos⁡A+cos⁡2A=1\cos A + \cos^2 A = 1 then sin⁡2A+sin⁡4A\sin^2 A + \sin^4 A is equal to:

Hunting for the angle

cos⁡θ+cos⁡2θ=1\cos\theta + \cos^2\theta = 1 has no table angle: cos θ works out to 5−12\frac{\sqrt{5} - 1}{2}. Trying 30°, 45° and 60° wastes the 36 seconds. Make the swap cos⁡θ=sin⁡2θ\cos\theta = \sin^2\theta instead.
Drill 4 more on substitution chains such as cos θ + cos²θ = 1

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)

  • sec θ + tan θ and sec θ − tan θ multiply to 1

    The reciprocal pairs

    (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1(cosec θ+cot⁡θ)(cosec θ−cot⁡θ)=1sec⁡θ+tan⁡θ=p⇒sin⁡θ=p2−1p2+11+sin⁡θ1−sin⁡θ=sec⁡θ+tan⁡θ(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1 \qquad (\text{cosec}\,\theta + \cot\theta)(\text{cosec}\,\theta - \cot\theta) = 1 \qquad \sec\theta + \tan\theta = p \Rightarrow \sin\theta = \frac{p^2 - 1}{p^2 + 1} \qquad \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} = \sec\theta + \tan\theta
  • Squaring a given sum

    Squared sums

    (sin⁡θ±cos⁡θ)2=1±2sin⁡θcos⁡θ(sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2=2x+1x=k⇒x2+1x2=k2−2tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ(\sin\theta \pm \cos\theta)^2 = 1 \pm 2\sin\theta\cos\theta \qquad (\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2 \qquad x + \frac{1}{x} = k \Rightarrow x^2 + \frac{1}{x^2} = k^2 - 2 \qquad \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta}
  • sin⁴θ + cos⁴θ and sin⁶θ + cos⁶θ

    Higher powers of sin and cos

    sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θsin⁡6θ+cos⁡6θ=1−3sin⁡2θcos⁡2θsin⁡4θ−cos⁡4θ=sin⁡2θ−cos⁡2θ\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta \qquad \sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta \qquad \sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta
  • Substitution chains such as cos θ + cos²θ = 1

    The two chains

    cos⁡θ+cos⁡2θ=1⇒cos⁡θ=sin⁡2θsin⁡θ+sin⁡2θ=1⇒sin⁡θ=cos⁡2θ\cos\theta + \cos^2\theta = 1 \Rightarrow \cos\theta = \sin^2\theta \qquad \sin\theta + \sin^2\theta = 1 \Rightarrow \sin\theta = \cos^2\theta

Watch out for (7)

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