Exercise 11.4 — Trigonometric Identities

Trigonometric identities and their applications.

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Exercise 11.4 — Trigonometric Identities

Exercise 11.4 is the algebraic heart of Chapter 11, Trigonometry, in Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus). It derives the three Pythagorean identities directly from the Pythagoras theorem, then uses them to evaluate expressions, simplify complicated fractions, and prove a wide range of "show that" statements that appear constantly in board exams.

This exercise has ten questions covering three broad skills: evaluating mixed trig expressions using identity substitution; proving identities by transforming the left-hand side step by step until it matches the right-hand side; and solving "if...then prove" problems where a given relationship between sec, tan, cosec, or cot leads to a hidden result.

Pythagorean Identities Evaluating Expressions Proving Identities If-Then Proofs
💡 Foundation fact: A trigonometric identity is an equation involving trig ratios that holds true for every value of the angle, not just specific ones. The three Pythagorean identities derived in this exercise are the most-used identities in the entire chapter — memorising them is non-negotiable for board exam success.

Deriving the Three Pythagorean Identities

Consider a right triangle ABC with ∠B = 90°. By the Pythagoras theorem, the three sides are related by:

BC² + AB² = AC²
A B C AB BC AC

Right triangle ABC with ∠B = 90°. AC is the hypotenuse, BC and AB are the legs.

Identity 1: sin²A + cos²A = 1

Dividing the Pythagoras equation throughout by AC² gives:

BC²/AC² + AB²/AC² = AC²/AC² (BC/AC)² + (AB/AC)² = 1 sin²A + cos²A = 1 (Identity 1)

Identity 2: sec²A − tan²A = 1

Dividing sin²A + cos²A = 1 throughout by cos²A gives:

sin²A/cos²A + cos²A/cos²A = 1/cos²A tan²A + 1 = sec²A sec²A − tan²A = 1 (Identity 2)

Identity 3: cosec²A − cot²A = 1

Dividing sin²A + cos²A = 1 throughout by sin²A gives:

sin²A/sin²A + cos²A/sin²A = 1/sin²A 1 + cot²A = cosec²A cosec²A − cot²A = 1 (Identity 3)
IdentityUseful Rearrangements
sin²A + cos²A = 1sin²A = 1−cos²A  |  cos²A = 1−sin²A  |  sinA = √(1−cos²A)  |  cosA = √(1−sin²A)
sec²A − tan²A = 1sec²A = 1+tan²A  |  tan²A = sec²A−1  |  secA = √(1+tan²A)  |  tanA = √(sec²A−1)
cosec²A − cot²A = 1cosec²A = 1+cot²A  |  cot²A = cosec²A−1  |  cosecA = √(1+cot²A)  |  cotA = √(cosec²A−1)

Alongside the three Pythagorean identities, every proof in this exercise also relies on the basic reciprocal and quotient identities:

sinθ·cosecθ = 1,   cosθ·secθ = 1,   tanθ·cotθ = 1
tanθ = sinθ/cosθ,   cotθ = cosθ/sinθ
💡 Memory trick: All three Pythagorean identities follow the same shape — "square of one ratio minus square of its paired ratio equals 1" — for the three pairs (sin,cos), (sec,tan), and (cosec,cot). Spotting which pair appears in a question instantly tells you which identity to reach for.
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Question 1 — Evaluate the Following Expressions

Each part of this question looks complicated at first glance, but converts into a clean numerical answer once everything is rewritten in terms of sinθ and cosθ, or once a Pythagorean identity is spotted hiding inside the algebra.

Part (i)
(1 + tanθ + secθ)(1 + cotθ − cosecθ)
= (1 + sinθ/cosθ + 1/cosθ)(1 + cosθ/sinθ − 1/sinθ) = [(cosθ+sinθ+1)/cosθ] × [(sinθ+cosθ−1)/sinθ] = [(cosθ+sinθ)²−1] / (cosθ·sinθ) [∵ (a+b)(a−b)=a²−b², with a=cosθ+sinθ, b=1] = [cos²θ+sin²θ+2sinθcosθ−1] / (cosθ·sinθ) = [1+2sinθcosθ−1] / (cosθ·sinθ) [∵ sin²θ+cos²θ=1] = 2sinθcosθ / (cosθ·sinθ) = 2
Answer: (1+tanθ+secθ)(1+cotθ−cosecθ) = 2
Part (ii)
(sinθ + cosθ)² + (sinθ − cosθ)²
= (sin²θ+cos²θ+2sinθcosθ) + (sin²θ+cos²θ−2sinθcosθ) [∵ (a±b)²=a²+b²±2ab] = 1 + 2sinθcosθ + 1 − 2sinθcosθ [∵ sin²θ+cos²θ=1] = 1 + 1 = 2
Answer: (sinθ+cosθ)² + (sinθ−cosθ)² = 2
Part (iii)
(sec²θ − 1)(cosec²θ − 1)
= (tan²θ)(cot²θ) [∵ sec²θ−1=tan²θ and cosec²θ−1=cot²θ] = (tanθ·cotθ)² [∵ tanθ·cotθ=1] = 1² = 1
Answer: (sec²θ−1)(cosec²θ−1) = 1
📌 Pattern across Question 1: Every part collapses cleanly to a small whole number once the Pythagorean identities and the algebraic shortcuts (a+b)(a−b)=a²−b² and (a±b)²=a²+b²±2ab are applied — recognising these algebra patterns is just as important as knowing the trig identities themselves.

Question 2 — Show That (cosecθ − cotθ)² = (1 − cosθ) ÷ (1 + cosθ)

Worked Solution
LHS: (cosecθ − cotθ)²
LHS = (1/sinθ − cosθ/sinθ)² = [(1−cosθ)/sinθ]² = (1−cosθ)² / sin²θ = (1−cosθ)² / (1−cos²θ) [∵ sin²θ=1−cos²θ] = (1−cosθ)(1−cosθ) / [(1−cosθ)(1+cosθ)] [∵ a²−b²=(a+b)(a−b)] = (1−cosθ) / (1+cosθ) = RHS
Hence proved: (cosecθ − cotθ)² = (1−cosθ) / (1+cosθ)

Question 3 — Show That (1+sinA) ÷ (1−sinA) = secA + tanA

Worked Solution
LHS: √[(1+sinA) / (1−sinA)]
Multiply numerator and denominator inside the root by (1+sinA): = √[(1+sinA)² / {(1−sinA)(1+sinA)}] = √[(1+sinA)² / (1−sin²A)] [∵ (a+b)(a−b)=a²−b²] = √[(1+sinA)² / cos²A] [∵ 1−sin²A=cos²A] = (1+sinA) / cosA = 1/cosA + sinA/cosA = secA + tanA = RHS
Hence proved: √[(1+sinA)/(1−sinA)] = secA + tanA
💡 Shared technique: Both Question 2 and Question 3 use the same trick — multiply numerator and denominator by a matching term to create a difference-of-squares (a+b)(a−b) pattern, which then converts neatly into sin²θ or cos²θ using a Pythagorean identity. This "rationalise, then substitute" approach is one of the most reusable techniques in the whole chapter.

Question 4 — Show That (1−tan²A) ÷ (cot²A−1) = tan²A

Worked Solution
LHS: (1 − tan²A) / (cot²A − 1)
LHS = (1−tan²A) / (1/tan²A − 1) [∵ cotA = 1/tanA] = (1−tan²A) / [(1−tan²A)/tan²A] = (1−tan²A) × [tan²A / (1−tan²A)] = tan²A = RHS
Hence proved: (1−tan²A) / (cot²A−1) = tan²A

Question 5 — Show That (1÷cosθ) − cosθ = tanθ · sinθ

Worked Solution
LHS: 1/cosθ − cosθ
LHS = (1 − cos²θ) / cosθ = sin²θ / cosθ [∵ 1−cos²θ=sin²θ] = (sinθ/cosθ) · sinθ = tanθ · sinθ = RHS
Hence proved: 1/cosθ − cosθ = tanθ · sinθ

Question 6 — Simplify secA(1 − sinA)(secA + tanA)

Worked Solution
= (secA − sinA·secA)(secA + tanA) = (secA − sinA/cosA)(secA + tanA) [∵ secA=1/cosA] = (secA − tanA)(secA + tanA) [∵ sinA/cosA = tanA] = sec²A − tan²A [∵ (a+b)(a−b)=a²−b²] = 1 (Pythagorean identity)
Answer: secA(1−sinA)(secA+tanA) simplifies to 1.

Question 7 — Show That (sinA+cosecA)² + (cosA+secA)² = 7 + tan²A + cot²A

Worked Solution
LHS: (sinA+cosecA)² + (cosA+secA)²
= sin²A+cosec²A+2sinA·cosecA + cos²A+sec²A+2cosA·secA [expand both squares] = (sin²A+cos²A) + cosec²A + sec²A + 2(sinA·cosecA) + 2(cosA·secA) = 1 + (1+cot²A) + (1+tan²A) + 2(1) + 2(1) [sinA·cosecA=1, cosA·secA=1] = 1 + 1 + cot²A + 1 + tan²A + 2 + 2 = 7 + tan²A + cot²A = RHS
Hence proved: (sinA+cosecA)² + (cosA+secA)² = 7 + tan²A + cot²A

Question 8 — Simplify (1−cosθ)(1+cosθ)(1+cot²θ)

Worked Solution
= (1−cos²θ)(1+cot²θ) [∵ (a+b)(a−b)=a²−b²] = (sin²θ)(cosec²θ) [∵ 1−cos²θ=sin²θ and 1+cot²θ=cosec²θ] = (sinθ·cosecθ)² = 1² = 1
Answer: (1−cosθ)(1+cosθ)(1+cot²θ) simplifies to 1.

Question 9 — If secθ + tanθ = p, Find secθ − tanθ

This question is a clever shortcut application of the Pythagorean identity sec²θ − tan²θ = 1, factored as a difference of squares.

Worked Solution
Given: secθ + tanθ = p
We know: sec²θ − tan²θ = 1 (secθ + tanθ)(secθ − tanθ) = 1 [∵ a²−b²=(a+b)(a−b)] p · (secθ − tanθ) = 1 [substituting the given value] secθ − tanθ = 1/p
Answer: If secθ + tanθ = p, then secθ − tanθ = 1/p.

Question 10 — If cosecθ + cotθ = k, Prove cosθ = (k²−1)÷(k²+1)

This is the most involved proof in the exercise, combining the same difference-of-squares trick from Question 9 with an extra step of solving for sinθ before converting to cosθ.

Worked Solution
Given: cosecθ + cotθ = k  ...(1)
We know: cosec²θ − cot²θ = 1 (cosecθ + cotθ)(cosecθ − cotθ) = 1 k · (cosecθ − cotθ) = 1 [using equation (1)] cosecθ − cotθ = 1/k  ...(2) Adding (1) and (2): 2cosecθ = k + 1/k = (k²+1)/k cosecθ = (k²+1) / 2k ⟹ sinθ = 2k / (k²+1) Now: cosθ = √(1 − sin²θ) cos²θ = 1 − [2k/(k²+1)]² = [(k²+1)² − 4k²] / (k²+1)² = [k⁴ + 2k² + 1 − 4k²] / (k²+1)² [expanding (k²+1)²] = [k⁴ − 2k² + 1] / (k²+1)² = (k²−1)² / (k²+1)² cosθ = (k²−1) / (k²+1) ✓ Hence proved
Hence proved: If cosecθ + cotθ = k, then cosθ = (k²−1) / (k²+1).
Watch the algebra in Q10: The step k⁴+2k²+1−4k² = k⁴−2k²+1 = (k²−1)² is a perfect-square factorisation that is easy to miscalculate under exam pressure — double-check this line carefully when reproducing the proof.

Common Mistakes to Avoid

  • Converting too early or too late: Most proofs in this exercise become solvable only after rewriting everything in terms of sinθ and cosθ — if you get stuck, that conversion is usually the missing step.
  • Misapplying (a+b)(a−b)=a²−b²: This algebraic identity appears in nearly every question (2, 3, 6, 8, 9, 10) — always look for a sum and a difference of the same two terms before assuming you can use it.
  • Forgetting which Pythagorean identity to use: sin²+cos²=1 pairs with (sinθ,cosθ); sec²−tan²=1 pairs with (secθ,tanθ); cosec²−cot²=1 pairs with (cosecθ,cotθ) — using the wrong pairing is a frequent source of stuck proofs.
  • Sign errors when expanding squares: (a−b)² = a² + b² − 2ab, not a² − b² — confusing this with the difference-of-squares formula is a very common slip, especially in Question 2 and Question 7.
  • Not simplifying the final answer fully: In Question 10, leaving cos²θ unsquare-rooted, or not factoring k⁴−2k²+1 as a perfect square, means the proof is incomplete even if all the algebra up to that point is correct.
  • Writing "1/p" as "p" by mistake: In Question 9, remember the final relationship is a reciprocal (secθ−tanθ = 1/p), not p itself — a careless final line can undo an otherwise correct solution.
High-risk exam trap: "If...then prove" questions like Q9 and Q10 are extremely popular for 3–4 mark board questions because they require recognising the difference-of-squares trick rather than memorising a fixed formula — practise spotting (a+b)(a−b)=1 patterns whenever a sum of two reciprocal-paired ratios is given.

Quick Reference — All Answers at a Glance

QuestionTopicKey Result
Q1(i)(1+tanθ+secθ)(1+cotθ−cosecθ)2
Q1(ii)(sinθ+cosθ)²+(sinθ−cosθ)²2
Q1(iii)(sec²θ−1)(cosec²θ−1)1
Q2(cosecθ−cotθ)²= (1−cosθ)/(1+cosθ), proved
Q3√[(1+sinA)/(1−sinA)]= secA+tanA, proved
Q4(1−tan²A)/(cot²A−1)= tan²A, proved
Q51/cosθ − cosθ= tanθ·sinθ, proved
Q6secA(1−sinA)(secA+tanA)1
Q7(sinA+cosecA)²+(cosA+secA)²= 7+tan²A+cot²A, proved
Q8(1−cosθ)(1+cosθ)(1+cot²θ)1
Q9secθ+tanθ=psecθ−tanθ = 1/p
Q10cosecθ+cotθ=kcosθ = (k²−1)/(k²+1), proved

What This Exercise Prepares You For

Exercise 11.4 is the most algebra-intensive exercise in the Trigonometry chapter, and mastering it pays off immediately in the final exercises of the chapter, which apply these same identities to heights and distances problems involving angles of elevation and depression. The standard-angle values from Exercise 11.2 and the complementary-angle identities from Exercise 11.3 are both used alongside the Pythagorean identities whenever a proof needs a specific numeric check.

The algebraic factoring skills practised throughout this exercise — especially (a+b)(a−b)=a²−b² and (a±b)²=a²+b²±2ab — also connect directly back to Algebraic Expressions and Factorisation, both of which provide the foundation for manipulating these identity proofs confidently.

📐 Board Exam Tip (CBSE, Telangana & AP): "Show that" and "if...then prove" identity questions from Exercise 11.4 are among the highest-weightage question types in the Trigonometry chapter, frequently worth 3–4 marks each. Practise writing out every algebraic step explicitly (including the identity being used at each line, as shown above) since partial proofs without clear justification often lose marks even when the final answer is correct.
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