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Question
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
Solution
`cos theta/(1 + sin theta)` = sec θ – tan θ
R.H.S. = sec θ – tan θ
= `1/cos theta - sin theta/cos theta`
= `(1 - sin theta)/costheta`
= `(1 - sin theta)/cos theta xx (1 + sin theta)/(1 + sin theta)`
= `(1 - sin^2 theta)/(cos theta(1 + sin theta)`
= `cos^2 theta/(cos theta(1 + sin theta))`
= `costheta/(1 + sintheta)`
L.H.S. = R.H.S.
∴ `cos theta/(1 + sin theta)` = sec θ – tan θ
Aliter:
L.H.S. = `cos theta/(1 - sin theta)` ...[conjugate (1 – sin θ)]
= `(cos theta(1 + sin theta))/((1 - sin theta)(1 + sin theta))`
= `(cos theta(1 + sin theta))/((1 - sin^2 theta))`
= `(cos theta (1 + sin theta))/(cos^2 theta)`
= `(1 + sin theta)/costheta`
L.H.S. = R.H.S.
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RELATED QUESTIONS
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
Prove that:
(cosec A – sin A) (sec A – cos A) sec2 A = tan A
What is the value of (1 + cot2 θ) sin2 θ?
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
`(sin A)/(1 + cos A) + (1 + cos A)/(sin A)` = 2 cosec A
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
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
If 1 + sin2θ = 3sinθ cosθ, then prove that tanθ = 1 or `1/2`.
