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प्रश्न
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
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उत्तर
In the given question, we need to prove `1 + cot^2 theta/(1 + cosec theta) = cosec theta`
Using `cot theta = cos theta/sin theta` and `cosec theta = 1/sin theta` We get
`1 + cot^2 theta/(1 + cosec theta) = (1 = cosec theta + cot^2 theta)/(1 + cosec theta)`
`= ((1 + 1/sin theta + cos^2 theta/sin^2 theta))/((1 + 1/sin theta))`
` = (((sin^2 theta + sin theta + cos^2 theta)/sin^2 theta))/(((sin theta + 1)/sin theta))`
Further, using the property `sin^2 theta + cos^2 theta = 1`
We get
`((sin^2 theta + sin theta + cos^2 theta)/sin^2 theta)/((sin theta + 1)/sin theta) = ((1 + sin theta)/sin^2 theta)/((sin theta + 1)/sin theta)`
`= (1 + sin theta/sin^2 theta)((sin theta)/(1 + sin theta))`
`= 1/sin theta`
`= cosec theta`
Hence proved.
संबंधित प्रश्न
If sinθ + sin2 θ = 1, prove that cos2 θ + cos4 θ = 1
Show that `sqrt((1+cosA)/(1-cosA)) = cosec A + cot A`
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Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
Prove the following identities:
`1/(secA + tanA) = secA - tanA`
Prove that:
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
Prove that:
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
Prove the following identity :
`(cotA + tanB)/(cotB + tanA) = cotAtanB`
If sinA + cosA = `sqrt(2)` , prove that sinAcosA = `1/2`
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
`(sin A)/(1 + cos A) + (1 + cos A)/(sin A)` = 2 cosec A
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
cos 45° = ?
