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प्रश्न
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
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उत्तर
We have to prove sec A(1 − sin A)(sec A + tan A) = 1
We know that sec2 A − tan2 A − 1
So,
sec A(1 − sin A)(sec A + tan A) = {sec A(1 − sin A)}(sec A + tan A)
= (sec A − sec A sin A)(sec A + tan A)
= `(sec A - 1/cos A sin A) (sec A + tan A)` ...`(∵ sec theta = 1/costheta)`
= `(sec A - sin A/cos A) (sec A + tan A)` ...`(∵ tan theta = sin theta/costheta)`
= (sec A − tan A)(sec A + tan A)
= sec2 A − tan2 A
= 1 = R.H.S. ... (∵ sec2 θ = 1 tan2 θ)
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(cot A + tan B)/(cot B + tan A) = cot A tan B`
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
Show that none of the following is an identity:
`tan^2 theta + sin theta = cos^2 theta`
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`
If sin θ = `11/61`, find the values of cos θ using trigonometric identity.
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
If `sec θ + tan θ = sqrt(3)`, complete the activity to find the value of sec θ – tan θ.
Activity:
`square = 1 + tan^2θ` ...[Fundamental trigonometric identity]
`square - tan^2θ = 1`
`(sec θ + tan θ) . (sec θ - tan θ) = square`
`sqrt(3) . (sec θ - tan θ) = 1`
`(sec θ - tan θ) = square`
