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प्रश्न
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
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उत्तर
L.H.S. = `(cotA + cosecA - 1)/(cotA - cosecA + 1)`
= `(cotA + cosecA - (cosec^2A - cot^2A))/(cotA - cosecA + 1)` ...[cosec2A – cot2A = 1]
= `(cotA + cosecA - [(cosecA - cotA)(cosecA + cotA)])/(cotA - cosecA + 1`
= `(cotA + cosecA[1 - cosecA + cotA])/(cotA - cosecA + 1)`
= cot A + cosec A
= `cosA/sinA + 1/sinA`
= `(1 + cosA)/sinA`
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संबंधित प्रश्न
Prove the following trigonometric identities
tan2 A + cot2 A = sec2 A cosec2 A − 2
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Choose the correct alternative:
1 + tan2 θ = ?
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
Prove the following identities.
tan4 θ + tan2 θ = sec4 θ – sec2 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.
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
