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प्रश्न
Prove the following identities:
`sqrt((1 - sinA)/(1 + sinA)) = cosA/(1 + sinA)`
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उत्तर
L.H.S. = `sqrt((1 - sinA)/(1 + sinA))`
= `sqrt((1 - sinA)/(1 + sinA) xx (1 + sinA)/(1 + sinA))`
= `sqrt((1 - sin^2A)/(1 + sinA)^2)`
= `sqrt(cos^2A/(1 + sinA)^2)`
= `cosA/(1 + sinA)` = R.H.S.
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संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.
Prove the following trigonometric identities.
tan2 θ − sin2 θ = tan2 θ sin2 θ
Prove the following trigonometric identities.
`sqrt((1 - cos A)/(1 + cos A)) = cosec A - cot A`
Prove the following identities:
`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`
Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
Prove the following identities: sec2 θ + cosec2 θ = sec2 θ cosec2 θ.
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
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.
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
