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CDS Mathematics · Trigonometric Ratios and Identities

Reciprocal Pairs: sec ± tan, cosec ± cot

Because sec²θ − tan²θ = 1, the numbers sec θ + tan θ and sec θ − tan θ are reciprocals of each other — and the same holds for cosec θ ± cot θ. One fact, and a whole family of questions collapses.

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Why this matters

Eighteen PYQs, all but one MODERATE, and CDS sets this family every year. They look like long fractions with ones scattered through them; each is solved in two lines by the reciprocal pair. It is the highest-return single identity in the chapter.

Concept 1 of 3: sec θ + tan θ and sec θ − tan θ are reciprocals

sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1 is a difference of squares, so it factors: (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1. Knowing one of the pair gives the other for free — and adding or subtracting the two recovers sec⁡θ\sec\theta and tan⁡θ\tan\theta separately.

Definition

  • (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1
  • (cosec⁡θ+cot⁡θ)(cosec⁡θ−cot⁡θ)=1(\operatorname{cosec}\theta + \cot\theta)(\operatorname{cosec}\theta - \cot\theta) = 1

If sec⁡θ+tan⁡θ=k\sec\theta + \tan\theta = k, then sec⁡θ−tan⁡θ=1k\sec\theta - \tan\theta = \dfrac1k, so

  • sec⁡θ=12(k+1k)\sec\theta = \dfrac12\left(k + \dfrac1k\right) and tan⁡θ=12(k−1k)\tan\theta = \dfrac12\left(k - \dfrac1k\right).

The cosecant pair works identically: cosec⁡θ\operatorname{cosec}\theta is half the sum, cot⁡θ\cot\theta half the difference.

The reciprocal pair

sec⁡θ+tan⁡θ=k  ⇒  sec⁡θ−tan⁡θ=1k,sec⁡θ=12(k+1k)\sec\theta + \tan\theta = k \;\Rightarrow\; \sec\theta - \tan\theta = \frac1k, \quad \sec\theta = \frac12\left(k + \frac1k\right)

Worked example

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

The same idea in a real exam question:

CDS · 2022 · CDS (I) 2022 — Elementary Mathematics · Q56Moderate

Example 1 · Trigonometric Ratios and Identities · Reciprocal Pairs — sec ± tan and cosec ± cot

If tan⁡θ+sec⁡θ=3\tan\theta + \sec\theta = 3, then what is the value of 3tan⁡θ+9sec⁡θ3\tan\theta + 9\sec\theta?

Half the sum gives sec, half the difference gives tan — not the other way

sec⁡θ=12(k+1k)\sec\theta = \dfrac12\left(k + \dfrac1k\right) is at least 11, as a secant must be. If your 'secant' comes out below 11, you have swapped the sum and the difference.
Drill 9 more on sec θ + tan θ and sec θ − tan θ are reciprocals

Concept 2 of 3: (1 + sin θ)/cos θ is sec θ + tan θ

Split the fraction: 1+sin⁡θcos⁡θ=1cos⁡θ+sin⁡θcos⁡θ=sec⁡θ+tan⁡θ\dfrac{1 + \sin\theta}{\cos\theta} = \dfrac{1}{\cos\theta} + \dfrac{\sin\theta}{\cos\theta} = \sec\theta + \tan\theta. Recognising that disguise is half the family, because the paper writes the reciprocal pair in sine-and-cosine form so you do not see it.

Definition

  • 1+sin⁡θcos⁡θ=sec⁡θ+tan⁡θ\dfrac{1 + \sin\theta}{\cos\theta} = \sec\theta + \tan\theta and 1−sin⁡θcos⁡θ=sec⁡θ−tan⁡θ\dfrac{1 - \sin\theta}{\cos\theta} = \sec\theta - \tan\theta.
  • 1+cos⁡θsin⁡θ=cosec⁡θ+cot⁡θ\dfrac{1 + \cos\theta}{\sin\theta} = \operatorname{cosec}\theta + \cot\theta and 1−cos⁡θsin⁡θ=cosec⁡θ−cot⁡θ\dfrac{1 - \cos\theta}{\sin\theta} = \operatorname{cosec}\theta - \cot\theta.
  • Under a root: 1−sin⁡θ1+sin⁡θ=1−sin⁡θ∣cos⁡θ∣\sqrt{\dfrac{1 - \sin\theta}{1 + \sin\theta}} = \dfrac{1 - \sin\theta}{|\cos\theta|}, which is sec⁡θ−tan⁡θ\sec\theta - \tan\theta when cos⁡θ>0\cos\theta > 0. Multiply inside by 1−sin⁡θ1 - \sin\theta top and bottom to see it.

The disguised pair

1±sin⁡θcos⁡θ=sec⁡θ±tan⁡θ,1±cos⁡θsin⁡θ=cosec⁡θ±cot⁡θ\frac{1 \pm \sin\theta}{\cos\theta} = \sec\theta \pm \tan\theta, \qquad \frac{1 \pm \cos\theta}{\sin\theta} = \operatorname{cosec}\theta \pm \cot\theta

Worked example

Simplify 1+cos⁡θ1−cos⁡θ\sqrt{\dfrac{1 + \cos\theta}{1 - \cos\theta}} for 0<θ<90∘0 < \theta < 90^\circ.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (I) 2020 — Elementary Mathematics · Q62Moderate

Example 2 · Trigonometric Ratios and Identities · Reciprocal Pairs — sec ± tan and cosec ± cot

What is sec⁡x−tan⁡xsec⁡x+tan⁡x\sqrt{\frac{\sec x - \tan x}{\sec x + \tan x}} equal to ?

A square root returns a modulus

cos⁡2θ=∣cos⁡θ∣\sqrt{\cos^2\theta} = |\cos\theta|. The simplification to sec⁡θ−tan⁡θ\sec\theta - \tan\theta is right only where cos⁡θ>0\cos\theta > 0. When the stem gives a range, check it; when it does not, the intended range is the first quadrant.
Drill 2 more on (1 + sin θ)/cos θ is sec θ + tan θ

Concept 3 of 3: Replacing the 1 in (tan θ + sec θ − 1)/(tan θ − sec θ + 1)

The standing CDS question is a fraction like tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1\dfrac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1}. The trick: the lonely 11 in the numerator is sec⁡2θ−tan⁡2θ\sec^2\theta - \tan^2\theta. Write it that way, factor, and the denominator appears as a factor of the numerator and cancels.

Definition

  • In the numerator replace −1-1 by tan⁡2θ−sec⁡2θ=(tan⁡θ−sec⁡θ)(tan⁡θ+sec⁡θ)\tan^2\theta - \sec^2\theta = (\tan\theta - \sec\theta)(\tan\theta + \sec\theta).
  • Then the numerator is (tan⁡θ+sec⁡θ)[1+tan⁡θ−sec⁡θ](\tan\theta + \sec\theta)\big[1 + \tan\theta - \sec\theta\big], whose bracket is the denominator.
  • So tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1=tan⁡θ+sec⁡θ=1+sin⁡θcos⁡θ\dfrac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1} = \tan\theta + \sec\theta = \dfrac{1 + \sin\theta}{\cos\theta}.

The same move works with cot⁡θ\cot\theta and cosec⁡θ\operatorname{cosec}\theta, and on the sine–cosine form sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1}, which equals sec⁡θ+tan⁡θ\sec\theta + \tan\theta too.

The standing result

tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1=sec⁡θ+tan⁡θ=1+sin⁡θcos⁡θ\frac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1} = \sec\theta + \tan\theta = \frac{1 + \sin\theta}{\cos\theta}

Worked example

Simplify cot⁡θ+cosec⁡θ−1cot⁡θ−cosec⁡θ+1\dfrac{\cot\theta + \operatorname{cosec}\theta - 1}{\cot\theta - \operatorname{cosec}\theta + 1}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (II) 2022 — Elementary Mathematics · Q54Moderate

Example 3 · Trigonometric Ratios and Identities · Reciprocal Pairs — sec ± tan and cosec ± cot

If x=1+sin⁡θcos⁡θx = \frac{1 + \sin\theta}{\cos\theta}, then what is tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1\frac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1} equal to ?

Cross-multiplying works, but costs three minutes

Every member of this family can be proved by clearing denominators and expanding, and every one of them then takes a page. Replacing the 11 takes two lines. If a fraction has a bare ±1\pm 1 beside tan⁡\tan and sec⁡\sec (or cot⁡\cot and cosec⁡\operatorname{cosec}), reach for the replacement first.
Drill 4 more on replacing the 1 in (tan θ + sec θ − 1)/(tan θ − sec θ + 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 (3)

  • sec θ + tan θ and sec θ − tan θ are reciprocals

    The reciprocal pair

    sec⁡θ+tan⁡θ=k  ⇒  sec⁡θ−tan⁡θ=1k,sec⁡θ=12(k+1k)\sec\theta + \tan\theta = k \;\Rightarrow\; \sec\theta - \tan\theta = \frac1k, \quad \sec\theta = \frac12\left(k + \frac1k\right)
  • (1 + sin θ)/cos θ is sec θ + tan θ

    The disguised pair

    1±sin⁡θcos⁡θ=sec⁡θ±tan⁡θ,1±cos⁡θsin⁡θ=cosec⁡θ±cot⁡θ\frac{1 \pm \sin\theta}{\cos\theta} = \sec\theta \pm \tan\theta, \qquad \frac{1 \pm \cos\theta}{\sin\theta} = \operatorname{cosec}\theta \pm \cot\theta
  • Replacing the 1 in (tan θ + sec θ − 1)/(tan θ − sec θ + 1)

    The standing result

    tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1=sec⁡θ+tan⁡θ=1+sin⁡θcos⁡θ\frac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1} = \sec\theta + \tan\theta = \frac{1 + \sin\theta}{\cos\theta}

Watch out for (3)

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