Advertisements
Advertisements
प्रश्न
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Advertisements
उत्तर
LHS = `cosA/(1 - tanA) + sin^2A/(sinA - cosA)`
= `cosA/(1 - sinA/cosA) + sin^2A/(sinA - cosA)`
= `cosA/((cosA - sinA)/(cosA)) + sin^2A/(sinA - cosA)`
= `cos^2A/((cosA - sinA)) - sin^2A/((cosA - sinA))`
= `(cos^2A - sin^2A)/(cosA - sinA) = ((cosA + sinA)(cosA - sinA))/((cosA - sinA))`
= (cosA + sinA)
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
sin2 A cot2 A + cos2 A tan2 A = 1
Prove the following trigonometric identities.
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
Prove the following trigonometric identities.
`(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove the following identities:
`1 - sin^2A/(1 + cosA) = cosA`
`(1-cos^2theta) sec^2 theta = tan^2 theta`
Prove the following identity :
`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`
If A = 60°, B = 30° verify that tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`.
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
