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प्रश्न
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
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उत्तर
LHS = `(sec A - 1)/(sec A + 1)`
= `(1/cos A - 1)/(1/cos A + 1)`
= `((1 - cos A)/cos A)/((1 + cos A)/cos A)`
= `(1 - cos A)/(1 + cos A)`
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities:
`(1 - cos^2 A) cosec^2 A = 1`
Prove the following trigonometric identities.
`sqrt((1 - cos theta)/(1 + cos theta)) = cosec theta - cot theta`
Prove the following trigonometric identities.
`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
`(1 + cot^2 theta ) sin^2 theta =1`
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Prove the following identity :
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
