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प्रश्न
Prove the following identities:
`sinA/(1 - cosA) - cotA = cosecA`
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उत्तर
`sinA/(1 - cosA) - cotA`
= `sinA/(1 - cosA) - cosA/sinA`
= `(sin^2A - cosA + cos^2A)/((1 - cosA)sinA)`
= `(1 - cosA)/((1 - cosA)sinA)`
= `1/sinA`
= cosec A
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संबंधित प्रश्न
Prove the following trigonometric identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
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 `( sin theta + cos theta ) = sqrt(2) , " prove that " cot theta = ( sqrt(2)+1)`.
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
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)
Prove that `(cos(90^circ - A))/(sin A) = (sin(90^circ - A))/(cos A)`.
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` ...`[cos theta = 1/sectheta]`
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
