Advertisements
Advertisements
प्रश्न
If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)
Advertisements
उत्तर
LHS = (b - c)(r - p) = `(b - asin^2θ - bcos^2θ)(p sin^2θ + qcos^2θ - p)`
= `[b(1 - cos^2θ) - asin^2θ][p(sin^2θ - 1) + q cos^2θ]`
⇒ LHS = `[(b - a)sin^2θ][(q - p)cos^2θ] = (b - a)(q - p)sin^2θcos^2θ`
RHS = `(c - a)(q- r) = (asin^2θ + bcos^2θ - a)(q - p sin^2θ - qcos^2θ)`
= `[(b - a)cos^2θ][(q - p)sin^2θ] = (b - a)(q - p)sin^2θ.cos^2θ`
Thus , (b - c)(r - p) = (c - a)(q - r)
APPEARS IN
संबंधित प्रश्न
If sinθ + sin2 θ = 1, prove that cos2 θ + cos4 θ = 1
Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
Prove that sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ.
(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
Prove that (sec θ + tan θ) (1 – sin θ) = cos θ
