if sin squared theta + b cos square theta equal to C . P sin square theta + cos square theta equal to r. prove that (b-c)(r-p)=(c-a)(q-r)
Answers
Question :-
if a.sin²θ + b.cos²θ = c and p.sin²θ + q.cos²θ = r , Prove that, (b-c)(r-p) = (c-a)(q-r) ?
Solution :-
Given that,
→ a.sin²θ + b.cos²θ = c
→ a.sin²θ + b.cos²θ = c * 1
Putting 1 = sin²A + cos²A in RHS now,
→ a.sin²θ + b.cos²θ = c * (sin²θ + cos²θ)
→ a.sin²θ + b.cos²θ = c.sin²θ + c.cos²θ
→ b.cos²θ - c.cos²θ = c.sin²θ - a.sin²θ
→ cos²θ (b - c) = sin²θ (c - a)
→ (b - c)/(c - a) = (sin²θ/cos²θ)
using (sinA/cosA) = tanA Now,
→ tan²θ = (b - c)/(c - a) ---------- Eqn(1)
Similarly,
→ p.sin²θ + q.cos²θ = r * 1
→ p.sin²θ + q.cos²θ = r * (sin²θ + cos²θ)
→ p.sin²θ + q.cos²θ = r.sin²θ + r.cos²θ
→ q.cos²θ - r.cos²θ = r.sin²θ - p.sin²θ
→ cos²θ (q - r) = sin²θ (r - p)
→ (q - r)/(r - p) = (sin²θ/cos²θ)
using (sinA/cosA) = tanA Now,
→ tan²θ = (q - r)/(r - p) ---------- Eqn(2)
From Eqn. (1) & (2) we get,
→ (b - c)/(c - a) = (q - r)/(r - p)
Cross - Multiply,
→(b - c) (r - p) = (c - a) (q - r) (Proved).
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