Advertisements
Advertisements
प्रश्न
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
Advertisements
उत्तर
We need to prove `cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos `
Solving the L.H.S, we get
`cos A/(1 - tan A) + sin A/(1 - cot A)`
= `cos A/(1 - sin A/cos A) + sin A/(1 - cos A/sin A)`
`= cos A/((cos A - sin A)/cos A) + sin A/((sin A - cos A)/sin A)`
`= cos^2 A/(cos A - sin A) + (sin^2 A)/(sin A - cos A)`
`= (cos^2 A - sin^2 A)/(cos A - sin A)`
`= ((cos A + sin A)(cos A - sin A))/(cos A - sin A)` [using `a^2 - b^2 = (a + b)(a -b)`]
= cos A + sin A
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities
(1 + cot2 A) sin2 A = 1
Prove the following trigonometric identities.
`sin theta/(1 - cos theta) = cosec theta + cot theta`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
`cosec theta (1+costheta)(cosectheta - cot theta )=1`
If x=a `cos^3 theta and y = b sin ^3 theta ," prove that " (x/a)^(2/3) + ( y/b)^(2/3) = 1.`
If x = a sin θ and y = bcos θ , write the value of`(b^2 x^2 + a^2 y^2)`
If cos A + cos2 A = 1, then sin2 A + sin4 A =
Prove the following identity :
( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
Prove that `(sin^2θ)/(cos θ) + cos θ = sec θ`.
