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प्रश्न
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
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उत्तर
LHS = `(sec A - 1)/(sec A + 1)`
= `(1/cos A - 1)/(1/cos A + 1)`
= `((1 - cos A)/cos A)/((1 + cos A)/cos A)`
= `(1 - cos A)/(1 + cos A)`
= RHS
Hence proved.
संबंधित प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Prove the following trigonometric identities.
`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`
Prove the following trigonometric identities.
tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
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.
Prove that (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B.
If 2sin2β − cos2β = 2, then β is ______.
(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.
(1 + sin A)(1 – sin A) is equal to ______.
