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प्रश्न
Prove the following identity :
`sec^2A + cosec^2A = sec^2Acosec^2A`
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उत्तर
LHS = `sec^2A + cosec^2A`
= `1/cos^2A + 1/sin^2A = (sin^2A + cos^2A)/(cos^2A.sin^2A)`
= `1/(cos^2A.sin^2A) = sec^2Acosec^2A` = RHS
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संबंधित प्रश्न
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
If x = asecθ + btanθ and y = atanθ + bsecθ , prove that `x^2 - y^2 = a^2 - b^2`
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
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)
If tan θ = `9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that `1/("cosec" theta - cot theta)` = cosec θ + cot θ
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ
