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प्रश्न
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
उत्तर
L.H.S. = cot θ + tan θ
= `costheta/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/(sintheta costheta)` ...[cos2 θ + sin2 θ = 1]
= `1/(sintheta costheta)`
= sec θ . cosec θ = R.H.S.
∴ cot θ + tan θ = sec θ cosec θ
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
Prove that:
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1
Prove the following identity :
`(cosA + sinA)^2 + (cosA - sinA)^2 = 2`
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 sin( 90° - θ ) sin θ cot θ = cos2θ.
Prove that `sqrt((1 + cos A)/(1 - cos A)) = (tan A + sin A)/(tan A. sin A)`
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 1.
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)
