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प्रश्न
Prove the following identities:
cosec A(1 + cos A) (cosec A – cot A) = 1
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उत्तर
L.H.S. = cosec A(1 + cos A) (cosecA – cot A)
= `1/(sin A)(1 + cos A)(1/(sin A) - (cos A)/(sin A))`
`((1-cos A)/sin A)`
`1/sin A(1+cos A)((1-cos A)/sin A)`
`= ((1+ cos A)(1-cos A))/sin^2 A`
Apply the identity (1 + cosA) (1 − cosA) = 1 − cos2A
`= (1-cos^2A)/sin^2A`
`= sin^2A/sin^2A = 1`
cscA(1 + cosA) (cscA − cotA) = 1 proved
संबंधित प्रश्न
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove the following identities:
`sinA/(1 - cosA) - cotA = cosecA`
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
Prove that: `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ) = tan θ`.
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
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)
Choose the correct alternative:
1 + cot2θ = ?
