Advertisements
Advertisements
प्रश्न
If cosθ = `5/13`, then find sinθ.
Advertisements
उत्तर
cosθ = `5/13`
`sin^2θ + cos^2θ = 1`
`sin^2θ + (5/13)^2 = 1`
`sin^2θ = (1 - 25)/169`
`sin^2θ = (169 - 25)/169`
`sin^2θ = 144/169`
sinθ = `sqrt(144/169)`
sinθ = `12/13`
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
Prove the following identities:
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
Prove the following identities:
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove the following identities:
`(sinA - cosA + 1)/(sinA + cosA - 1) = cosA/(1 - sinA)`
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
Show that none of the following is an identity:
`cos^2theta + cos theta = 1`
Show that none of the following is an identity:
`tan^2 theta + sin theta = cos^2 theta`
Write the value of `(sin^2 theta 1/(1+tan^2 theta))`.
Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:
sin θ × cosec θ = ______
What is the value of (1 + cot2 θ) sin2 θ?
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
Prove the following identity :
`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Prove that `tan^3 θ/( 1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ) = sec θ. cosec θ - 2 sin θ cos θ.`
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
If `sec θ = 41/40`, then find values of sin θ, cot θ, cosec θ.
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
