Advertisements
Advertisements
प्रश्न
If `cos θ = 24/25`, then sin θ = ?
Advertisements
उत्तर
`cos θ = 24/25` ...[Given]
We know that,
sin2θ + cos2θ = 1
∴ `sin^2θ + (24/25)^2 = 1`
∴ `sin^2θ + 576/625 = 1`
∴ `sin^2θ = 1 - 576/625`
∴ `sin^2θ = (625 - 576)/625`
∴ `sin^2θ = 49/625`
∴ `sin θ = 7/25` ...[Taking square root of both sides]
APPEARS IN
संबंधित प्रश्न
Express the ratios cos A, tan A and sec A in terms of sin A.
Prove the following identities:
`("cosec" theta + cot theta)/("cosec" theta - cot theta) = ("cosec" theta + cot theta )^2 = 1 + 2 cot^2 theta + 2 "cosec" theta cot theta`
If `(x/a sin theta - y/b cos theta) = 1` and `(x/a cos theta + y/b sin theta) = 1`, prove that `(x^2/a^2 + y^2/b^2) = 2`.
If `cos theta = 7/25 , "write the value of" ( tan theta + cot theta).`
Prove that:
Sin4θ - cos4θ = 1 - 2cos2θ
\[\frac{x^2 - 1}{2x}\] is equal to
Prove the following identity :
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
Find the value of sin 30° + cos 60°.
Prove that: 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0.
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
Prove that `(sin θ + "cosec" θ)/(sin θ) = 2 + cot^2θ`.
Prove that 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0.
