Advertisements
Advertisements
प्रश्न
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
Advertisements
उत्तर
`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.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 - tan^2 A)/(cot^2 A -1) = tan^2 A`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = sec A + tan A`
Prove the following identities:
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
Prove that:
`1/(cosA + sinA - 1) + 1/(cosA + sinA + 1) = cosecA + secA`
` tan^2 theta - 1/( cos^2 theta )=-1`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
cot θ . tan θ = ?
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` ...`[cos theta = 1/sectheta]`
