Advertisements
Advertisements
Question
If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)
Advertisements
Solution
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
RELATED QUESTIONS
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
`cot^2 theta - 1/(sin^2 theta ) = -1`a
Write the value of `(cot^2 theta - 1/(sin^2 theta))`.
If `sec theta = x ,"write the value of tan" theta`.
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
Prove the following identity :
`sqrt((secq - 1)/(secq + 1)) + sqrt((secq + 1)/(secq - 1))` = 2 cosesq
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
Prove that `"cosec" θ xx sqrt(1 - cos^2θ) = 1`.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
