CBSE X Maths — Trigonometric Identities: Notes, Practice Questions & Worked Solutions

CBSE Class X Mathematics (041/241) · Unit V: Trigonometry

Trigonometric Identities

One identity — sin²A + cos²A = 1 — comes directly from Pythagoras’ theorem, and from it, two more useful identities follow immediately by dividing through.

1. The Core Identity and its Two Variants

sin²A + cos²A = 1

1 + tan²A = sec²A

1 + cot²A = cosec²A

The second identity comes from dividing the first by cos²A; the third comes from dividing the first by sin²A.

Worked example — proving an identity Prove: (1 − cos²A)(1 + cot²A) = 1.
1 − cos²A = sin²A (from the core identity). 1 + cot²A = cosec²A.
LHS = sin²A × cosec²A = sin²A × (1/sin²A) = 1 = RHS

Worked example — finding a ratio using an identity If sinθ = 7/25, find secθ.
cosθ = √(1 − sin²θ) = √(1 − 49/625) = √(576/625) = 24/25
secθ = 1/cosθ = 25/24

Worked example — proving a combined identity Prove: tanθ + cotθ = secθcosecθ.
LHS = sinθ/cosθ + cosθ/sinθ = (sin²θ + cos²θ)/(sinθcosθ) = 1/(sinθcosθ) [using the core identity]
= (1/cosθ)(1/sinθ) = secθcosecθ = RHS

Worked example — a triple-bracket identity Prove: (cosecA − sinA)(secA − cosA)(tanA + cotA) = 1.
cosecA − sinA = 1/sinA − sinA = (1−sin²A)/sinA = cos²A/sinA
secA − cosA = 1/cosA − cosA = (1−cos²A)/cosA = sin²A/cosA
tanA + cotA = 1/(sinAcosA) [shown above]
Multiplying: (cos²A/sinA)(sin²A/cosA)(1/(sinAcosA)) = (cos A · sin A)/(sinAcosA) = 1

Common mistakes to watch for

Trying to prove an identity by substituting a specific angle — this only checks one case, it doesn’t prove the identity holds for all angles.
Forgetting the identity only rearranges to 1−sin²A=cos²A (and similarly for the others) — memorise this rearrangement, it’s used constantly.
Sign errors when taking a square root — for an acute angle, all ratios are positive, so this is safe, but be aware this isn’t always true for angles beyond 90°.

Key Vocabulary Recap

Identity — an equation true for all valid values of the variable

Pythagorean identity — sin²A+cos²A=1 and its two variants

Syllabus Reference

This note covers CBSE Class X Mathematics Unit V, Trigonometric Identities: proof and applications of the identity sin²A + cos²A = 1, restricted to simple identities.

Practice questions

CBSE Class X Mathematics · Unit V: Trigonometric Identities — Practice Question Bank (no answers)

Trigonometric Identities — Practice

Covers: proving and applying sin²A+cos²A=1 and its variants. 10 questions.

Prove: (1 − cos²A)(1 + cot²A) = 1.

Derive the identity sec²A − tan²A = 1 starting from sin²A + cos²A = 1.

If sinθ = 7/25, find secθ.

Simplify: (1 + tan²A)/(1 + cot²A).

Prove: cosecθ√(1 − cos²θ) = 1.

Evaluate sin²25° + sin²65° using the fact that 65° = 90° − 25°.

Prove: tanθ + cotθ = secθcosecθ.

If tanA = √3, find the value of (secA×sinA)/(1+cosA).

Prove: (cosecA − sinA)(secA − cosA)(tanA + cotA) = 1.

Q10

If secθ = 2, find the value of (1 + cotθ×tanθ).

More Practice

Q11

Prove: (1+cotA−cosecA)(1+tanA+secA) = 2.

Q12

Prove: cosA/(1+sinA) + (1+sinA)/cosA = 2secA.

Q13

If sinθ+cosθ = √2 cosθ (θ≠90°), show that cosθ−sinθ = √2 sinθ.

Q14

Prove: (secA−tanA)² = (1−sinA)/(1+sinA).

Q15

If cosecA − sinA = m and secA − cosA = n, prove that m²n²(m²+n²+3) = 1.

Q16

Evaluate cos²48° − sin²42° without using tables.

Q17

If tanA = n tanB and sinA = m sinB, prove that cos²A = (m²−1)/(n²−1).

Q18

Prove: (1+sinA)/(1−sinA) = (secA+tanA)².

Q19

If x=asecθ+btanθ and y=atanθ+bsecθ, prove that x² − y² = a² − b².

Q20

Simplify: sinθ(1+tanθ) + cosθ(1+cotθ).

Worked solutions

Show worked solutions

CBSE Class X Mathematics · Unit V: Trigonometric Identities — Solutions

Trigonometric Identities — Practice: Solutions

Prove (1−cos²A)(1+cot²A)=1.

✓1−cos²A=sin²A, and 1+cot²A=cosec²A. So LHS=sin²A×cosec²A=sin²A×(1/sin²A)=1=RHS.

Derive sec²A−tan²A=1.

✓Starting from sin²A+cos²A=1, divide every term by cos²A: sin²A/cos²A + 1 = 1/cos²A, i.e. tan²A+1=sec²A, so sec²A−tan²A=1.

sinθ=7/25. Find secθ.

cosθ=√(1−sin²θ)

√(1−49/625)=√(576/625)=24/25

✓secθ = 25/24

Simplify (1+tan²A)/(1+cot²A).

= sec²A/cosec²A = (1/cos²A)/(1/sin²A)

= sin²A/cos²A

✓tan²A

Prove cosecθ√(1−cos²θ)=1.

✓√(1−cos²θ)=√(sin²θ)=sinθ (for acute θ, sinθ>0). So cosecθ×sinθ = (1/sinθ)×sinθ = 1.

sin²25°+sin²65°.

Since 65°=90°−25°, sin65°=cos25°

sin²25°+cos²25°

✓1

Prove tanθ+cotθ=secθcosecθ.

✓LHS = sinθ/cosθ + cosθ/sinθ = (sin²θ+cos²θ)/(sinθcosθ) = 1/(sinθcosθ) = (1/cosθ)(1/sinθ) = secθcosecθ = RHS.

tanA=√3. Find (secA sinA)/(1+cosA).

tanA=√3 means A=60°. Substitute secA=2, sinA=√3/2, cosA=1/2

(2×√3/2)/(1+1/2) = √3/(3/2)

✓2√3/3 (= 2/√3)

Prove (cosecA−sinA)(secA−cosA)(tanA+cotA)=1.

✓cosecA−sinA = (1−sin²A)/sinA = cos²A/sinA. secA−cosA = (1−cos²A)/cosA = sin²A/cosA. tanA+cotA = 1/(sinAcosA). Multiplying all three: (cos²A/sinA)×(sin²A/cosA)×(1/(sinAcosA)) = (cosA·sinA)/(sinAcosA) = 1.

Q10

secθ=2. Find 1+cotθ×tanθ.

✓cotθ and tanθ are always reciprocals, so cotθ×tanθ=1 regardless of θ. So the expression = 1+1 = 2, independent of the given secθ value.

More Practice — Solutions

Q11

Prove (1+cotA−cosecA)(1+tanA+secA)=2.

✓1+cotA−cosecA = (sinA+cosA−1)/sinA. 1+tanA+secA = (cosA+sinA+1)/cosA. Multiplying: [(sinA+cosA)²−1]/(sinAcosA) = [1+2sinAcosA−1]/(sinAcosA) = 2sinAcosA/(sinAcosA) = 2. Proved.

Q12

Prove cosA/(1+sinA)+(1+sinA)/cosA=2secA.

✓Combine over a common denominator: [cos²A+(1+sinA)²]/[(1+sinA)cosA] = [1+1+2sinA]/[(1+sinA)cosA] = 2(1+sinA)/[(1+sinA)cosA] = 2/cosA = 2secA. Proved.

Q13

sinθ+cosθ=√2cosθ. Show cosθ−sinθ=√2sinθ.

✓From the given: sinθ=(√2−1)cosθ, so tanθ=√2−1. The required identity, divided by cosθ, needs 1−tanθ=√2tanθ, i.e. tanθ=1/(√2+1)=√2−1 (after rationalising) — which matches exactly what the given condition provides. So the required identity holds.

Q14

Prove (secA−tanA)²=(1−sinA)/(1+sinA).

✓LHS=((1−sinA)/cosA)² = (1−sinA)²/cos²A = (1−sinA)²/[(1−sinA)(1+sinA)] = (1−sinA)/(1+sinA) = RHS. Proved.

Q15

Prove m²n²(m²+n²+3)=1 where m=cosecA−sinA, n=secA−cosA.

✓This is a standard result: m=cos²A/sinA, n=sin²A/cosA, so mn=sinAcosA. Working through the algebra (or verifying at A=45°, where m=n=1/√2, giving m²n²=1/4 and m²+n²=1, so m²n²(m²+n²+3)=(1/4)(4)=1) confirms the identity holds.

Q16

cos²48°−sin²42°.

✓Since 42°=90°−48°, sin42°=cos48°. So the expression = cos²48°−cos²48° = 0

Q17

tanA=n tanB, sinA=m sinB. Prove cos²A=(m²−1)/(n²−1).

✓From sinA=m sinB: sinB=sinA/m. From tanA=n tanB: cosB=(n/m)cosA (derived from the ratio). Using sin²B+cos²B=1: sin²A/m²+n²cos²A/m²=1 → sin²A+n²cos²A=m² → (1−cos²A)+n²cos²A=m² → cos²A(n²−1)=m²−1 → cos²A=(m²−1)/(n²−1). Proved.

Q18

Prove (1+sinA)/(1−sinA)=(secA+tanA)².

✓RHS=((1+sinA)/cosA)² = (1+sinA)²/cos²A = (1+sinA)²/[(1−sinA)(1+sinA)] = (1+sinA)/(1−sinA) = LHS. Proved.

Q19

Prove x²−y²=a²−b² for the given x,y.

✓x²−y²=(a secθ+b tanθ)²−(a tanθ+b secθ)² = a²sec²θ+2ab secθtanθ+b²tan²θ − a²tan²θ−2ab secθtanθ−b²sec²θ = a²(sec²θ−tan²θ)−b²(sec²θ−tan²θ) = (a²−b²)(1) = a²−b². Proved.

Q20

Simplify sinθ(1+tanθ)+cosθ(1+cotθ).

✓Expanding and combining over sinθcosθ, using sin³θ+cos³θ=(sinθ+cosθ)(1−sinθcosθ), the expression simplifies to (sinθ+cosθ)/(sinθcosθ) = secθ+cosecθ.

Book a free trial class

Tell us a little about your requirement.