Solutions — Pythagorean and Reciprocal Identities

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1. Q1 — Reciprocal Functions

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   The reciprocal functions are: cosecθ = 1/sinθ, secθ = 1/cosθ, cotθ = cosθ/sinθ.

   With sinθ = 5/13 and cosθ = 12/13:

   (a) cosecθ = 1/(5/13) = 13/5

   (b) secθ = 1/(12/13) = 13/12

   (c) tanθ = (5/13)/(12/13) = 5/13 × 13/12 = 5/12

   (d) cotθ = 1/tanθ = 12/5 = 12/5

   Verification: sin²θ + cos²θ = 25/169 + 144/169 = 169/169 = 1 ✓

2. Q2 — Evaluating Using Identities

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   Given tanθ = 3/4, use the identity sec²θ = 1 + tan²θ:

   sec²θ = 1 + (3/4)² = 1 + 9/16 = 25/16

   Since θ is in Q1, secθ > 0: secθ = 5/4, so cosθ = 4/5.

   Then sinθ = tanθ × cosθ = (3/4)(4/5) = 3/5.

   sinθ = 3/5,   cosθ = 4/5,   secθ = 5/4

   Verification: sin²θ + cos²θ = 9/25 + 16/25 = 1 ✓

3. Q3 — Simplifying Trigonometric Expressions

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   Convert each expression to sin and cos and simplify.

   (a) sinθ · cotθ = sinθ × (cosθ/sinθ)

   = cosθ    (sinθ cancels) = cosθ

   (b) cos²θ · sec²θ − 1 = cos²θ × (1/cos²θ) − 1

   = 1 − 1 = 0

   (c) (tanθ · cosecθ) / secθ

   = (sinθ/cosθ) × (1/sinθ) ÷ (1/cosθ)

   = (1/cosθ) × cosθ = 1

4. Q4 — Pythagorean Substitution

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   (a) (1 − sin²θ)/cosθ

   Apply sin²θ + cos²θ = 1 ⇒ 1 − sin²θ = cos²θ

   = cos²θ/cosθ = cosθ

   (b) sec²θ − tan²θ

   From 1 + tan²θ = sec²θ ⇒ sec²θ − tan²θ = 1

   (c) (cosec²θ − 1)/cosec²θ

   From 1 + cot²θ = cosec²θ ⇒ cosec²θ − 1 = cot²θ

   = cot²θ/cosec²θ = (cos²θ/sin²θ) × sin²θ = cos²θ

5. Q5 — Proving an Identity (Basic)

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   Prove: tanθ + cotθ = secθ · cosecθ.

   Work on the LHS. Convert to sin and cos to find a common denominator:

   LHS = sinθ/cosθ + cosθ/sinθ

   = sin²θ/(sinθ cosθ) + cos²θ/(sinθ cosθ)   [LCD = sinθ cosθ]

   = (sin²θ + cos²θ) / (sinθ cosθ)

   = 1 / (sinθ cosθ)   [since sin²θ + cos²θ = 1]

   = (1/sinθ)(1/cosθ)

   = cosecθ · secθ = secθ · cosecθ = RHS ✓

6. Q6 — Proving an Identity (Intermediate)

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   Prove: (sinθ + cosθ)² = 1 + 2 sinθ cosθ.

   Work on the LHS. Expand using (a+b)² = a² + 2ab + b²:

   LHS = sin²θ + 2 sinθ cosθ + cos²θ

   = (sin²θ + cos²θ) + 2 sinθ cosθ

   = 1 + 2 sinθ cosθ   [Pythagorean identity]

   = RHS ✓

7. Q7 — Finding All Ratios Given One

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   Given cosθ = −2/3, θ in Q3 (π < θ < 3π/2).

   In Q3: sin < 0, cos < 0, tan > 0.

   Finding sinθ:

   sin²θ = 1 − cos²θ = 1 − 4/9 = 5/9

   |sinθ| = √5/3. Since Q3, sinθ = −√5/3

   Finding tanθ:

   tanθ = sinθ/cosθ = (−√5/3)/(−2/3) = √5/2. Q3 confirms positive. tanθ = √5/2

   Finding cosecθ:

   cosecθ = 1/sinθ = 1/(−√5/3) = −3/√5 = −3√5/5 (rationalised)

8. Q8 — Proving an Identity (Factorising)

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   Prove: (1 − cosθ)(1 + cosθ) = sin²θ.

   Work on the LHS. This is the difference of two squares (a−b)(a+b) = a²−b² with a=1, b=cosθ:

   LHS = 1² − cos²θ = 1 − cos²θ

   = sin²θ   [since sin²θ + cos²θ = 1 ⇒ sin²θ = 1 − cos²θ]

   = RHS ✓

9. Q9 — Complex Identity Proof

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   Prove: sin⁴θ − cos⁴θ = sin²θ − cos²θ.

   Work on the LHS. Let u = sin²θ and v = cos²θ to see the structure clearly:

   LHS = u² − v² = (u+v)(u−v)   [difference of two squares]

   = (sin²θ + cos²θ)(sin²θ − cos²θ)

   = 1 × (sin²θ − cos²θ)   [Pythagorean identity]

   = sin²θ − cos²θ = RHS ✓

10. Q10 — Proving a Cosec/Cot Identity

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    Prove: (cosecθ − cotθ)(cosecθ + cotθ) = 1.

    Work on the LHS. Recognise the difference of squares pattern (a−b)(a+b) = a²−b²:

    LHS = cosec²θ − cot²θ

    By the Pythagorean identity 1 + cot²θ = cosec²θ:

    cosec²θ − cot²θ = 1

    = RHS ✓

    Alternative method (converting to sin/cos):

    LHS = (1/sinθ − cosθ/sinθ)(1/sinθ + cosθ/sinθ)

    = [(1−cosθ)/sinθ] × [(1+cosθ)/sinθ]

    = (1−cosθ)(1+cosθ)/sin²θ

    = (1−cos²θ)/sin²θ

    = sin²θ/sin²θ = 1 = RHS ✓