Question

If cot 0 + cos 0 = p and coto - cos 0 = q, then the
value of p2 - q2 is

$(1) 2\sqrt{pq}$
$(2)4 \sqrt{pq}$
$(3)2pq$
$(4)4pq$
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Answer 1

Given :

• cot θ + cos θ = p  ____equation (1)
• cot θ - cos θ = q   ____equation (2)

To find :

• p² - q²

option are:

(1) 2 √(pq)

(2) 4 √(pq)

(3) 2 pq

(4) 4 pq

Solution :

→ p² - q²

putting given values of p and q

→ p² - q²= (cot θ + cos θ)² - (cot θ - cos θ)²

using algebraic identity , ( a + b )² = a² + b² + 2 a b and ( a - b )² = a² + b² - 2 a b

→ p² - q² = (cot²θ + cos²θ + 2 cot θ cos) - (cot²θ + cos²θ - 2 cot θ cos θ)

→ p² - q² = cot²θ + cos²θ + 2 cot θ cos - cot²θ - cos²θ + 2 cot θ cos θ

→ p² - q² = 4 cot cos θ

→ ( p² - q²) / 4 = cot cos θ____equation (3)

Now,

Multiplying equation (1) and (2)

→ (cot θ + cos θ) (cot θ - cos θ) = p q

using algebraic identity, a² - b² = ( a + b ) ( a - b )

→ cot²θ - cos²θ = p q

putting cot²θ = cos²θ / sin²θ

→ (cos²θ/sin²θ) - cos²θ = p q

Taking LCM in LHS

→ ( cos²θ - sin²θ cos²θ ) / sin²θ = p q

Taking cos²θ common in numerator of LHS

→ [cos²θ ( 1 - sin²θ ) ] / sin²θ = p q

using trigonometric identity, 1 - sin²θ = cos²θ

→ ( cos²θ cos²θ ) / sin²θ = p q

→ cot²θ cos²θ = p q

→ ( cot θ cos θ )² = p q

→ cot θ cos θ = √(p q)

using equation (3)

→ ( p² - q² ) / 4 = √ (p q)

→ p² - q² = 4 √(p q)

therefore,

• p² - q² = 4 √( p q )

Option (2) is correct .