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प्रश्न
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
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उत्तर
R.H.S. = `(1 - sinA)/(1 + sinA)`
= `(1 - 1/(cosecA))/(1 + 1/(cosecA))`
= `(cosecA - 1)/(cosecA + 1)`
= `(cosecA - 1)/(cosecA + 1) xx (cosecA + 1)/(cosecA + 1)`
= `(cosec^2A - 1)/(cosecA + 1)^2 = cot^2A/(cosecA + 1)^2` ...(∵ cosec2 A – 1 = cot2 A)
= L.H.S.
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संबंधित प्रश्न
Prove the following trigonometric identities.
`tan A/(1 + tan^2 A)^2 + cot A/((1 + cot^2 A)) = sin A cos A`
If sin θ + cos θ = x, prove that `sin^6 theta + cos^6 theta = (4- 3(x^2 - 1)^2)/4`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
Prove the following identities:
`(sec theta + tan theta)/(sec theta - tan theta) = (sec theta + tan theta)^2 = 1 + 2 tan^2 theta + 2 sec theta tan theta`
If `sec theta + tan theta = p,` prove that
(i)`sec theta = 1/2 ( p+1/p) (ii) tan theta = 1/2 ( p- 1/p) (iii) sin theta = (p^2 -1)/(p^2+1)`
If `sec theta + tan theta = x," find the value of " sec theta`
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
Prove the following identity :
`sqrt((1 + cosA)/(1 - cosA)) = cosecA + cotA`
Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan 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.
