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प्रश्न
Prove the following identities:
`1/(secA + tanA) = secA - tanA`
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उत्तर १
L.H.S. = `1/(secA + tanA)`
= `1/(1/cosA + sinA/cosA)`
= `1/((1 + sinA)/cosA)`
= `cosA/(1 + sinA) xx (1 - sinA)/(1 + sinA)`
= `(cosA(1 - sinA))/((1)^2 - sin^2A)`
= `(cosA(1 - sinA))/cos^2A`
= `1/cosA - sinA/cosA`
= sec A – tan A
L.H.S. = R.H.S.
Hence proved.
उत्तर २
L.H.S = `1/(secA + tanA)`
= `((secA - tanA))/((secA + tanA)(secA - tanA))` ...((Multiply Num. and Deno. by sec A – tan A)
= `(secA - tanA)/(sec^2A - tan^2A)`
= `(secA - tanA)/1` ...[∵ sec2 A – tan2 A = 1]
= sec A – tan A
= R.H.S.
संबंधित प्रश्न
Prove the following trigonometric identities:
`(1 - cos^2 A) cosec^2 A = 1`
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
Prove the following identities:
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
If `cosA/cosB = m` and `cosA/sinB = n`, show that : (m2 + n2) cos2 B = n2.
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.
