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प्रश्न
Prove the following identities:
`(secA - tanA)/(secA + tanA) = 1 - 2secAtanA + 2tan^2A`
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उत्तर
L.H.S. = `(secA - tanA)/(secA + tanA)`
= `(secA - tanA)/(secA + tanA) xx (secA - tanA)/(secA - tanA)`
= `(secA - tanA)^2/(sec^2A - tan^2A)`
= `(sec^2A + tan^2A - 2secAtanA)/1`
= 1 + tan2 A + tan2 A – 2 sec A tan A
= 1 – 2 sec A tan A + 2 tan2 A = R.H.S.
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संबंधित प्रश्न
If secθ + tanθ = p, show that `(p^{2}-1)/(p^{2}+1)=\sin \theta`
Prove the following identities:
`sqrt((1 - sinA)/(1 + sinA)) = cosA/(1 + sinA)`
Prove the following identities:
`(cos theta "cosec" theta - sin theta sec theta )/(cos theta + sin theta) = "cosec" theta - sec theta`
Write the value of `( 1- sin ^2 theta ) sec^2 theta.`
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Which is not correct formula?
If tan θ × A = sin θ, then A = ?
If 3 sin θ = 4 cos θ, then sec θ = ?
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
