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प्रश्न
If cosθ = `5/13`, then find sinθ.
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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`
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संबंधित प्रश्न
Prove the following trigonometric identities:
`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `
Prove the following trigonometric identities
(1 + cot2 A) sin2 A = 1
Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
Prove the following identities:
sec2 A . cosec2 A = tan2 A + cot2 A + 2
Prove the following identities:
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove the following identities:
`cosA/(1 + sinA) + tanA = secA`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
Prove that:
tan (55° + x) = cot (35° – x)
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Prove that: `(1 + cot^2 θ/(1 + cosec θ)) = cosec θ`.
sin2θ + sin2(90 – θ) = ?
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
Prove that `sqrt((1 + cos A)/(1 - cos A)) = "cosec" A + cot A`.
