Trigonometry for SSC CHSL

Many SSC CHSL aspirants skip trigonometry because it looks hard. That is an expensive mistake. The questions are Class 10 level and follow a handful of patterns. A candidate who knows the value table, three identities and the complementary-angle rule can answer most of them in under 40 seconds, faster than a typical percentage question.

Expect trigonometry in both Tier 1 and Tier 2 maths, usually as value questions, identity simplifications and one heights-and-distances problem. The exact count varies by shift, so use recent papers to judge the mix.

Ratios, and why one ratio gives all the others

In a right triangle, for an acute angle θ:

• sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent.
• cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ.
• tan θ = sin θ/cos θ.

The ratios depend only on the angle, not on the size of the triangle, because all right triangles with the same angle θ are similar. So if you know one ratio, draw a triangle with those two sides, find the third side by Pythagoras, and read off everything else.

The value table

A way to rebuild the sin row in the exam hall: write √0/2, √1/2, √2/2, √3/2, √4/2. That gives 0, 1/2, 1/√2, √3/2 and 1. The cos row is the same list backwards. tan is sin ÷ cos.

The values come from two triangles. Half an equilateral triangle has sides in the ratio 1 : √3 : 2 (angles 30°, 60°, 90°). A right isosceles triangle has sides 1 : 1 : √2 (angles 45°, 45°, 90°). These two triangles also solve most heights and distances questions directly.

The three identities

1. sin²θ + cos²θ = 1
2. 1 + tan²θ = sec²θ
3. 1 + cot²θ = cosec²θ

Why: in a right triangle, opposite² + adjacent² = hypotenuse². Divide by hypotenuse² to get identity 1. Divide by adjacent² to get identity 2, and by opposite² to get identity 3.

The most-used exam consequence is (sec θ + tan θ)(sec θ − tan θ) = 1, which is identity 2 rearranged. Similarly (cosec θ + cot θ)(cosec θ − cot θ) = 1. If a question gives sec θ + tan θ, it is almost certainly asking you to use this.

Complementary angles

• sin(90° − θ) = cos θ and cos(90° − θ) = sin θ
• tan(90° − θ) = cot θ and sec(90° − θ) = cosec θ

Why: the two acute angles of a right triangle add to 90°. The side opposite one is adjacent to the other, so the sine of one is the cosine of the other.

Whenever two angles in a question add to 90° (25° and 65°, 10° and 80°), convert one into the co-ratio of the other and look for cancellation.

The value-putting shortcut

If a question asks you to simplify an identity that holds for all θ, put in a convenient angle and test the options. For example, (sin θ + cos θ)² + (sin θ − cos θ)² equals what? Put θ = 0°: (0 + 1)² + (0 − 1)² = 2. The answer is 2.

Choose 0°, 30°, 45° or 60°, and avoid angles where a term is undefined, such as tan 90°. If two options give the same value, try a second angle.

Heights and distances

Use tan θ = height ÷ horizontal distance. Draw the figure, mark the right angle at the foot of the tower or tree, and use the 1 : √3 : 2 or 1 : 1 : √2 triangle.

Six worked questions

Q1. If tan θ = 8/15, find sin θ + cos θ.
Sides 8 and 15 give hypotenuse 17 (an 8-15-17 triplet). sin θ + cos θ = 8/17 + 15/17 = 23/17.

Q2. Find the value of 4 sin²60° + 3 tan²30° − 8 sin 45° cos 45°.
4 × 3/4 = 3. 3 × 1/3 = 1. 8 × (1/√2) × (1/√2) = 4. Total = 3 + 1 − 4 = 0.

Q3. If sec θ + tan θ = 3, find sec θ.
Since (sec θ + tan θ)(sec θ − tan θ) = 1, sec θ − tan θ = 1/3. Add the two equations: 2 sec θ = 10/3, so sec θ = 5/3. Check: tan θ = 4/3 and 25/9 − 16/9 = 1.

Q4. Find sin 25° sec 65° + cos 25° cosec 65°.
sec 65° = 1/cos 65° = 1/sin 25°, so the first term is 1. cosec 65° = 1/sin 65° = 1/cos 25°, so the second term is 1. Total = 2.

Q5. A 6 m pole casts a shadow 2√3 m long. Find the angle of elevation of the sun.
tan θ = 6/(2√3) = √3, so θ = 60°.

Q6. From a point on the ground, the angle of elevation of the top of a tower is 30°. After walking 20 m towards the tower, it becomes 60°. Find the height.
Let the height be h. Distance at 30° = h√3; at 60° = h/√3. The difference is 20: h√3 − h/√3 = 2h/√3 = 20, so h = 10√3 m, about 17.3 m.

Common mistakes

Practice

1. If cos θ = 5/13, find sin θ and tan θ.
2. Find 2 sin 30° + 3 tan 45°.
3. If cosec θ − cot θ = 1/4, find cosec θ.
4. Find tan 10° × tan 20° × tan 70° × tan 80°.
5. From a point 50 m from a building, the angle of elevation of its top is 60°. Find the building's height.
6. Find cos²45° + sin²30°.
7. A 10 m ladder leans against a wall, making 60° with the ground. How high up the wall does it reach?
8. If sin θ = cos θ for an acute angle θ, find θ and the value of tan θ + cot θ.

Answers:

1. sin θ = 12/13, tan θ = 12/5. A 5-12-13 triangle.
2. 4. 2 × 1/2 + 3 × 1.
3. 17/8. cosec θ + cot θ = 4, so 2 cosec θ = 4 + 1/4 = 17/4.
4. 1. tan 10° × tan 80° = 1 and tan 20° × tan 70° = 1.
5. 50√3 m, about 86.6 m. tan 60° = h/50.
6. 3/4. 1/2 + 1/4.
7. 5√3 m, about 8.66 m. Height = 10 × sin 60°.
8. θ = 45°; the value is 2. tan 45° = cot 45° = 1.

What to do next

• Rebuild the value table from the √0/2 … √4/2 pattern until it takes under 20 seconds.
• Solve 15 identity questions, trying value-putting on each to check your algebra.
• Do 10 heights and distances questions, drawing the triangle every time.
• Revise geometry for right triangles and algebra for the x + 1/x style that trigonometry questions borrow.