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प्रश्न
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
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उत्तर
LHS = `1 - (cos^2 θ)/(1 + sin θ)`
= `1 - (1 - sin^2 θ)/(1 + sin θ)`
= `1 - ((1 - sin θ)(1 + sin θ))/(1 + sin θ)`
= 1 - ( 1 - sin θ )
= 1 - 1 + sin θ
= sin θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`cot theta - tan theta = (2 cos^2 theta - 1)/(sin theta cos theta)`
`(sin theta+1-cos theta)/(cos theta-1+sin theta) = (1+ sin theta)/(cos theta)`
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
Prove that `tan^3 θ/( 1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ) = sec θ. cosec θ - 2 sin θ cos 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.
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
If 2sin2θ – cos2θ = 2, then find the value of θ.
