मराठी
महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

If Cosθ = 5 13 , Then Find Sinθ.

Advertisements
Advertisements

प्रश्न

If cosθ = `5/13`, then find sinθ. 

बेरीज
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`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2018-2019 (July) Set 1

APPEARS IN

संबंधित प्रश्‍न

Show that `sqrt((1-cos A)/(1 + cos A)) = sinA/(1 + cosA)`


Prove the following trigonometric identities

(1 + cot2 A) sin2 A = 1


Prove the following trigonometric identities.

`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`


Prove the following trigonometric identities.

`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`


Prove the following trigonometric identities.

`(1 + cos A)/sin A = sin A/(1 - cos A)`


Prove the following trigonometric identities.

`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`


Prove the following trigonometric identities.

`(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`


Prove the following identities:

`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`


Prove the following identities:

(1 + cot A – cosec A)(1 + tan A + sec A) = 2


What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]


Prove the following identity : 

`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`


Verify that the points A(–2, 2), B(2, 2) and C(2, 7) are the vertices of a right-angled triangle. 


Evaluate:
`(tan 65°)/(cot 25°)`


Prove that: 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0.


Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.


Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`


tan θ cosec2 θ – tan θ is equal to


`sin θ = 1/2`, then θ = ?


Prove that `(cos(90^circ - A))/(sin A) = (sin(90^circ - A))/(cos A)`.


Prove that sin6A + cos6A = 1 – 3sin2A . cos2A.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×