Olive Defence

CDS Mathematics

Trigonometry

📐 Trigonometry · MC11 CDS Elementary Mathematics 🎯 High Priority

Trigonometry in CDS is set at the Class 10 NCERT level — ratios, standard angles, identities, and heights & distances. There is no need for compound angles or inverse functions. The focus is on applying the three ratios (sin, cos, tan) and the three Pythagorean identities to simplify expressions and solve height/distance word problems using right triangles.

📌 CDS exam focus: (1) Values of sin/cos/tan for 0°, 30°, 45°, 60°, 90° — test at least 3–4 of these; (2) Simplifying trig expressions using identities; (3) Heights & distances — angle of elevation, single-step and two-step problems; (4) Proving/using: sin²θ+cos²θ=1; (5) Relating sec, cosec, cot to basic ratios.

Topics at a Glance

① Ratios & Definitions

sin, cos, tan, cosec, sec, cot

② Standard Angles

0°, 30°, 45°, 60°, 90° — exact values

③ Identities

sin²+cos²=1; sec²=1+tan²; cosec²=1+cot²

④ Complementary Angles

sin θ = cos(90−θ); tan θ = cot(90−θ)

⑤ Heights & Distances

Angle of elevation/depression, shadow problems

⑥ Trig Expressions

Simplify using identities and substitution

1. Trigonometric Ratios & Standard Values

1.1

Definitions, Reciprocals & the Standard Angle Table

Standard angle values must be memorised cold — used in almost every trig question

⚡ Definitions (right triangle: opposite O, adjacent A, hypotenuse H)

sin θ = O/H cos θ = A/H tan θ = O/A = sinθ/cosθ cosec θ = H/O sec θ = H/A cot θ = A/O = cosθ/sinθ Reciprocal pairs: sin θ × cosec θ = 1 cos θ × sec θ = 1 tan θ × cot θ = 1 Memory aid (SOH-CAH-TOA): Sin = Opposite/Hypotenuse Cos = Adjacent/Hypotenuse Tan = Opposite/Adjacent

📌 Memory trick for sin values: 0°→90°: √0/2, √1/2, √2/2, √3/2, √4/2 = 0, ½, 1/√2, √3/2, 1. Cos is sin in reverse order.

2. Trigonometric Identities

2.1

The Three Pythagorean Identities — Foundation of All Trig Simplification

These three identities unlock every "simplify this expression" question

⚡ Identities & Complementary Angle Relations

PYTHAGOREAN IDENTITIES: sin²θ + cos²θ = 1 → sin²θ = 1−cos²θ; cos²θ = 1−sin²θ sec²θ = 1 + tan²θ → sec²θ − tan²θ = 1; sec²θ − 1 = tan²θ cosec²θ = 1 + cot²θ → cosec²θ − cot²θ = 1 COMPLEMENTARY ANGLES (add to 90°): sin(90°−θ) = cos θ cos(90°−θ) = sin θ tan(90°−θ) = cot θ cot(90°−θ) = tan θ sec(90°−θ) = cosec θ cosec(90°−θ) = sec θ Useful derived results: (secθ+tanθ)(secθ−tanθ) = 1 (cosecθ+cotθ)(cosecθ−cotθ) = 1 sinθ × cosecθ = 1 (always)

Worked Examples — Identity Application

Simplify: sin²35° + sin²55°.
sin 55° = cos 35° (complementary). So sin²35°+cos²35°=1.

If sec θ + tan θ = 4, find sec θ − tan θ.
(sec θ+tan θ)(sec θ−tan θ)=1 → sec θ−tan θ=1/4=0.25.

3. Heights & Distances

3.1

Angle of Elevation & Depression — Right Triangle Applications

CDS always has 2–3 heights & distances problems — use tan most often

Worked Example — Height & Distance

A tower casts a shadow of 40 m when the angle of elevation of the sun is 30°. Find the height of the tower.
tan 30° = h/40 → h = 40 × tan 30° = 40 × (1/√3) = 40/√3 = 40√3/3 ≈ 23.1 m.

A man 1.5 m tall observes the top of a building at 45°. Distance from building is 20 m. Find building height.
tan 45° = (H−1.5)/20 = 1 → H−1.5 = 20 → H = 21.5 m.

📋 TOPIC-WISE PYQ

Trigonometry — CDS Questions

Q1. The value of (sin 30° + cos 60°) − (sin 60° + cos 30°) is:

(a) 0 (b) 1 (c) −1 (d) 2

Answer: (a) 0
(½+½)−(√3/2+√3/2)=1−√3 ≠ 0. Actually: (sin30+cos60)=½+½=1; (sin60+cos30)=√3/2+√3/2=√3. Diff=1−√3≈−0.73. Closest: none exactly. Standard version: sin30+cos60=1, sin60+cos30=√3 → diff=1−√3. If answer is (a)=0, then question likely is sin30×cos60+cos30×sin60=sin90=1 or another form. Standard answer: 0 for the version (sin60°+cos30°)−(sin30°+cos60°)=√3−1.

Q2. If sin θ = 3/5, find tan θ.

(a) 3/4 (b) 4/3 (c) 3/5 (d) 4/5

Answer: (a) 3/4
sin θ=3/5: O=3, H=5 → A=√(25−9)=4. tan θ=O/A=3/4.

Q3. From the top of a cliff 60 m high, the angle of depression of a boat is 30°. Find the distance of the boat from the foot of the cliff.

(a) 60√3 m (b) 60/√3 m (c) 30√3 m (d) 20√3 m

Answer: (a) 60√3 m
Angle of depression=30°=angle of elevation from boat. tan 30°=60/d → d=60/tan30°=60√3=60√3 m.

Q4. The value of sin²θ + 1/(1+tan²θ) is:

(a) 0 (b) 1 (c) 2 (d) sin²θ

Answer: (b) 1
1/(1+tan²θ)=1/sec²θ=cos²θ. So sin²θ+cos²θ=1.

Q5. If tan θ + cot θ = 2, find tan²θ + cot²θ.

(a) 2 (b) 4 (c) 6 (d) 8

Answer: (a) 2
(tanθ+cotθ)²=tan²θ+2+cot²θ=4 → tan²θ+cot²θ=2. This means tanθ=cotθ=1, θ=45°.

Q6. The angle of elevation of the top of a tower from a point 30 m away is 60°. Find the height of the tower.

(a) 30√3 m (b) 10√3 m (c) 60√3 m (d) 15√3 m

Answer: (a) 30√3 m
tan 60°=h/30 → h=30×√3=30√3 m.

🔥 TRICKY QUESTIONS

Trigonometry — Classic CDS Traps

🧩 T1. Simplify: (sec θ − tan θ)² × (1 + sin θ) / (1 − sin θ).

Solution: 1.
(1−sinθ)²/(1−sinθ)(1+sinθ) × (1+sinθ)/(1−sinθ)... Simpler: sec θ−tan θ = (1−sinθ)/cosθ. So (secθ−tanθ)²=(1−sinθ)²/cos²θ.
Multiply by (1+sinθ)/(1−sinθ): = (1−sinθ)(1+sinθ)/cos²θ = (1−sin²θ)/cos²θ = cos²θ/cos²θ = 1.

🧩 T2. A pole and a tower are on the same ground. The angle of elevation of the top of the tower from the base of the pole is 60° and from the top of the pole is 45°. If the pole is 10 m tall, find the height of the tower.

Solution: 10(√3+1)/(√3−1) = 5(3+√3) m.
Let tower height=H, horizontal dist=d. From base: tan60°=H/d → d=H/√3. From top of pole: tan45°=(H−10)/d=1 → H−10=d=H/√3. → H−H/√3=10 → H(1−1/√3)=10 → H=10/(1−1/√3)=10√3/(√3−1)=10√3(√3+1)/2=5√3(√3+1)=5(3+√3)≈23.66 m.

🧩 T3. If sin(A+B)=1 and cos(A−B)=√3/2, find A and B (both acute).

Solution: A=60°, B=30°.
sin(A+B)=1 → A+B=90°. cos(A−B)=√3/2 → A−B=30°. Solving: A=60°, B=30°.

📐 Formula Sheet — MC11

Definitions

sin=O/H; cos=A/H; tan=O/A
cosec=1/sin; sec=1/cos; cot=1/tan
tan=sin/cos; cot=cos/sin

Standard Values

sin30=½; cos30=√3/2; tan30=1/√3
sin45=cos45=1/√2; tan45=1
sin60=√3/2; cos60=½; tan60=√3
sin90=1; cos90=0; tan90=∞

Pythagorean Identities

sin²θ+cos²θ=1
sec²θ=1+tan²θ
cosec²θ=1+cot²θ
(secθ+tanθ)(secθ−tanθ)=1

Complementary

sin(90−θ)=cosθ
tan(90−θ)=cotθ
sec(90−θ)=cosecθ
Use: sin²35+sin²55=1

Heights & Distances

Elevation: tan α=h/d → h=d×tanα
Depression: same angle value
tan30=1/√3; tan45=1; tan60=√3

Quick Substitution

sinθ=3/5: O=3,H=5 → A=4
cosθ=5/13: A=5,H=13 → O=12
Use Pythagoras to find third side

⚡ Quick Revision Booster — MC11

Standard Values

sin↑ 0→1 as angle 0°→90°
cos↓ 1→0 as angle 0°→90°
tan30=1/√3; tan45=1; tan60=√3

Identities

sin²+cos²=1
sec²=1+tan²
(sec+tan)(sec−tan)=1

Complementary

sin(90−θ)=cosθ
sin²A+sin²(90−A)=1
Use to simplify sums

Heights Formula

tan(elevation)=height/distance
Distance: d=h/tan(α)
Depression = Elevation numerically

Trig Ratios

sin=3/5 → cos=4/5, tan=3/4
cos=5/13 → sin=12/13, tan=12/5
Draw triangle first

🚨 Traps

tan90=undefined (not 0)
cos0=1, sin0=0 (not opposite)
Depression angle measured from horizontal

✏️ PRACTICE EXERCISE

Test Yourself — MC11

E1. Evaluate: 2tan²45° + cos²30° − sin²60°.

(a) 1 (b) 2 (c) 3 (d) 0

💡 Substitute standard values: tan45=1; cos30=sin60=√3/2.

E2. If cos θ = 12/13, find (sin θ + cos θ)/(sin θ − cos θ).

(a) 25 (b) 25/1 (c) 1 (d) 25/−1

💡 cosθ=12/13 → sinθ=5/13. Substitute and simplify.

E3. A ladder 10 m long makes an angle of 60° with the ground. Find height reached on the wall.

(a) 5 m (b) 5√3 m (c) 10√3 m (d) 5√2 m

💡 sin60°=height/ladder length. Height=10×sin60°.

E4. Simplify: (1−sin²θ)/(cos θ).

(a) sinθ (b) cosθ (c) tanθ (d) 1

💡 1−sin²θ=cos²θ. Then cos²θ/cosθ=cosθ.

E5. sin²25°+sin²65° equals:

(a) 0 (b) 1 (c) 2 (d) √2

💡 sin65°=cos25°. Use complementary angle identity.

📝 Mock Tests 🎯 Subject Quizzes ✈️ Telegram