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SSC CGL Maths · Trigonometric Ratios and Identities

Ratios in a Right Triangle

Given one ratio or two sides, draw the right triangle, find the third side by Pythagoras, and read off any other ratio.

All Trigonometric Ratios and Identities notes

Why this matters

59 questions in the bank, the most of any page in this chapter, and 52 of them are EASY. 20 are from 2025, more than from either earlier year. Fifteen are a fraction in sin θ and cos θ with tan θ given, which one division turns into a single line.

Concept 1 of 3: Ratios from the sides of a right triangle

Before any ratio, decide which side is which. The hypotenuse never changes, but the opposite and adjacent sides swap when you move from one acute angle to the other.

Definition

  • The hypotenuse is opposite the right angle. In a triangle PQR right-angled at R, it is PQ, the side that does not touch R.
  • The side opposite angle A is adjacent to the other acute angle. That is why sin⁡A=cos⁡C\sin A = \cos C when B is the right angle.
  • Find a missing side with Pythagoras: leg2+leg2=hypotenuse2\text{leg}^2 + \text{leg}^2 = \text{hypotenuse}^2.
  • Learn the common triples so the third side is instant: 3, 4, 5 · 5, 12, 13 · 8, 15, 17 · 7, 24, 25 · 20, 21, 29 · 9, 40, 41. Any multiple works too (6, 8, 10).
  • A side from an angle: opposite =hypotenuse×sin⁡θ= \text{hypotenuse} \times \sin\theta, adjacent =hypotenuse×cos⁡θ= \text{hypotenuse} \times \cos\theta, opposite =adjacent×tan⁡θ= \text{adjacent} \times \tan\theta.

Sides of a right triangle

a2+b2=c2opposite=hypotenuse×sin⁡θadjacent=hypotenuse×cos⁡θa^2 + b^2 = c^2 \qquad \text{opposite} = \text{hypotenuse} \times \sin\theta \qquad \text{adjacent} = \text{hypotenuse} \times \cos\theta

Worked example

Triangle ABC is right-angled at B, with AB = 8 cm and AC = 17 cm. Find sin⁡C\sin C and tan⁡A\tan A.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 11 Sep 2024, Shift 3 · Q74Easy

Example 1 · Trigonometric Ratios and Identities · Ratios in a Right Triangle

Consider a triangle PQR, right angled at R, in which PQ = 29 units, QR = 21 units and ∠PQR=θ\angle PQR = \theta. Find the value of cos⁡2θ−sin⁡2θ\cos^2\theta - \sin^2\theta.

Using the longest printed number as a leg

Find the right angle first. The side that does not touch it is the hypotenuse, whatever order the sides are printed in. Subtract squares to get a leg; add squares only to get the hypotenuse.

Opposite and adjacent belong to one angle

In a right triangle, the side opposite one acute angle is adjacent to the other. Rename the sides every time the question moves to a different angle.
Drill 4 more on ratios from the sides of a right triangle

Concept 2 of 3: One ratio gives the other five

A ratio is a fraction of two sides, so you can draw a triangle with exactly those sides. Pythagoras gives the third side, and then every other ratio is a reading from the picture.

Definition

  • tan⁡θ=ab\tan\theta = \frac{a}{b}: opposite a, adjacent b, hypotenuse a2+b2\sqrt{a^2 + b^2}.
  • sin⁡θ=ac\sin\theta = \frac{a}{c}: opposite a, hypotenuse c, adjacent c2−a2\sqrt{c^2 - a^2}. For cos⁡θ\cos\theta, swap opposite and adjacent.
  • sec, cosec and cot: turn them over first. sec⁡θ=2920\sec\theta = \frac{29}{20} means hypotenuse 29, adjacent 20.
  • Letters work the same way: sin⁡A=mn\sin A = \frac{m}{n} gives cos⁡A=n2−m2n\cos A = \frac{\sqrt{n^2 - m^2}}{n}. For sin⁡A=2x1+x2\sin A = \frac{2x}{1 + x^2} the third side is 1−x21 - x^2, because (1+x2)2−(2x)2=(1−x2)2(1 + x^2)^2 - (2x)^2 = (1 - x^2)^2.
  • Fast route: spot the triple. cot⁡θ=247\cot\theta = \frac{24}{7} is the 7, 24, 25 triangle, so sin⁡θ=725\sin\theta = \frac{7}{25} with no square root to work out.
  • Limit: the triangle gives positive values, so it answers only for an acute angle. If the angle is in another quadrant, the signs change (Complementary and Allied Angles).

From tan θ = a/b

sin⁡θ=aa2+b2cos⁡θ=ba2+b2\sin\theta = \frac{a}{\sqrt{a^2 + b^2}} \qquad \cos\theta = \frac{b}{\sqrt{a^2 + b^2}}

Worked example

If 40tan⁡θ=940\tan\theta = 9 and θ is acute, find cos⁡θ−sin⁡θ\cos\theta - \sin\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 3 · Q68Easy

Example 2 · Trigonometric Ratios and Identities · Ratios in a Right Triangle

If sin⁡A=110\sin A = \frac{1}{\sqrt{10}}, then find the value of tan⁡A+sec⁡A\tan A + \sec A.

Using the triangle outside the first quadrant

If the angle lies between 90° and 360°, the triangle gives only the size of each ratio. In the second quadrant, for example, cos and tan are negative, so the sign must be fixed by the quadrant.

Rejecting a correct answer that looks different

8113\frac{8}{\sqrt{113}} and 8113113\frac{8\sqrt{113}}{113} are the same number. If no option matches your form, rationalise or un-rationalise before deciding you are wrong.
Drill 38 more on one ratio gives the other five

Concept 3 of 3: Dividing by cos θ when tan θ is given

In a fraction like 3sin⁡θ+2cos⁡θ3sin⁡θ−2cos⁡θ\frac{3\sin\theta + 2\cos\theta}{3\sin\theta - 2\cos\theta}, every term carries exactly one sin or one cos. Divide top and bottom by cos⁡θ\cos\theta and each sin becomes tan while each cos becomes 1, so the given tan goes straight in.

Definition

  • asin⁡θ+bcos⁡θcsin⁡θ+dcos⁡θ=atan⁡θ+bctan⁡θ+d\frac{a\sin\theta + b\cos\theta}{c\sin\theta + d\cos\theta} = \frac{a\tan\theta + b}{c\tan\theta + d}.
  • If cot is given, divide by sin⁡θ\sin\theta instead, and the fraction becomes one in cot⁡θ\cot\theta.
  • Fast route: tan⁡θ=25\tan\theta = \frac{2}{5} means you may put sin⁡θ=2\sin\theta = 2 and cos⁡θ=5\cos\theta = 5 straight into the fraction. The common scale cancels, so there is no hypotenuse to find.
  • Backwards: if the fraction is given equal to a number, cross-multiply, collect the sin terms on one side and the cos terms on the other, and read off tan.
  • Limit: every term must have the same degree. A lone number, or a tan mixed with sin and cos, breaks the cancellation. Then find sin and cos from the triangle and substitute.

Divide through by cos θ

asin⁡θ+bcos⁡θcsin⁡θ+dcos⁡θ=atan⁡θ+bctan⁡θ+d\frac{a\sin\theta + b\cos\theta}{c\sin\theta + d\cos\theta} = \frac{a\tan\theta + b}{c\tan\theta + d}

Worked example

If 4tan⁡θ=94\tan\theta = 9, find 8sin⁡θ−6cos⁡θ4sin⁡θ+6cos⁡θ\frac{8\sin\theta - 6\cos\theta}{4\sin\theta + 6\cos\theta}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 10 Sep 2024, Shift 3 · Q74Easy

Example 3 · Trigonometric Ratios and Identities · Ratios in a Right Triangle

If 7tan⁡θ=37\tan\theta = 3, and θ\theta is an acute angle, then 5sin⁡θ−cos⁡θ5sin⁡θ+2cos⁡θ\frac{5\sin\theta - \cos\theta}{5\sin\theta + 2\cos\theta} is equal to:

Dividing when a constant term is present

sin⁡θ+1cos⁡θ\frac{\sin\theta + 1}{\cos\theta} does not become a fraction in tan alone, because the 1 turns into sec⁡θ\sec\theta. The trick needs every term to be a single sin or cos.

Getting tan after dividing by sin

Dividing by sin⁡θ\sin\theta turns cos⁡θ\cos\theta into cot⁡θ\cot\theta, not tan. Divide by the ratio that leaves the one you were given.
Drill 14 more on dividing by cos θ when tan θ is given

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)

  • Ratios from the sides of a right triangle

    Sides of a right triangle

    a2+b2=c2opposite=hypotenuse×sin⁡θadjacent=hypotenuse×cos⁡θa^2 + b^2 = c^2 \qquad \text{opposite} = \text{hypotenuse} \times \sin\theta \qquad \text{adjacent} = \text{hypotenuse} \times \cos\theta
  • One ratio gives the other five

    From tan θ = a/b

    sin⁡θ=aa2+b2cos⁡θ=ba2+b2\sin\theta = \frac{a}{\sqrt{a^2 + b^2}} \qquad \cos\theta = \frac{b}{\sqrt{a^2 + b^2}}
  • Dividing by cos θ when tan θ is given

    Divide through by cos θ

    asin⁡θ+bcos⁡θcsin⁡θ+dcos⁡θ=atan⁡θ+bctan⁡θ+d\frac{a\sin\theta + b\cos\theta}{c\sin\theta + d\cos\theta} = \frac{a\tan\theta + b}{c\tan\theta + d}

Watch out for (6)

Test yourself on Trigonometric Ratios and Identities

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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