Advertisements
Advertisements
प्रश्न
Prove the following trigonometric identities.
`(tan A + tan B)/(cot A + cot B) = tan A tan B`
Advertisements
उत्तर
We have to prove `(tan A + tan B)/(cot A + cot B) = tan A tan B`
Now
`(tan A + tan B)/(cot A + cot B) = (tan A + tan B)/(1/tan A + 1/tanB)`
`= (tan A + tan B)/((tan B + tan A)/(tan A tan B))`
= tan A tan B
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
Prove the following identities:
`sqrt((1 - sinA)/(1 + sinA)) = cosA/(1 + sinA)`
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2
If `cosA/cosB = m` and `cosA/sinB = n`, show that : (m2 + n2) cos2 B = n2.
Prove the following identities:
`(1 - 2sin^2A)^2/(cos^4A - sin^4A) = 2cos^2A - 1`
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
`cot^2 theta - 1/(sin^2 theta ) = -1`a
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
Write the value of tan10° tan 20° tan 70° tan 80° .
What is the value of (1 + cot2 θ) sin2 θ?
If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
Prove that `sin A/(sec A + tan A - 1) + cos A/("cosec" A + cot A - 1) = 1`.
cos 45° = ?
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S. = `square`
= `cos^2θ xx square` ...`[1 + tan^2θ = square]`
= `(cos θ xx square)^2`
= 12
= 1
= R.H.S.
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
Prove that `sqrt((1 + cos A)/(1 - cos A)) = "cosec" A + cot A`.
