Advertisements
Advertisements
प्रश्न
If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`
Advertisements
उत्तर
Given that, tan A = n tan B and sin A = m sin B.
`=> n = tanA/tanB` and `m = sinA/sinB`
∴ `(m^2 - 1)/(n^2 - 1) = ((sinA/sinB)^2 - 1)/((tanA/tanB)^2 - 1)`
= `(sin^2A/sin^2B - 1/1)/(tan^2A/(tan^2B) - 1)`
= `((sin^2A - sin^2B).tan^2B)/(sin^2B.(tan^2A - tan^2B))`
= `((sin^2A - sin^2B)/tan^2B)/((tan^2A - tan^2B)/sin^2B)`
= `((sin^2A - sin^2B)sin^2B)/((sin^2A/cos^2A-sin^2B/cos^2B)cos^2Bsin^2B)`
= `(sin^2A - sin^2B)/(((sin^2A.cos^2B - sin^2B.cos^2A)/(cos^2A.cos^2B)) cos^2B)`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A.cos^2B - sin^2B.cos^2A)`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A(1 - sin^2B) - sin^2B (1 - sin^2A))`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A - sin^2A.sin^2B - sin^2B + sin^2B.sin^2A)`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A -sin^2B)`
= cos2 A
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(tan^3 theta)/(1 + tan^2 theta) + (cot^3 theta)/(1 + cot^2 theta) = sec theta cosec theta - 2 sin theta cos theta`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.
If tan α = n tan β, sin α = m sin β, prove that cos2 α = `(m^2 - 1)/(n^2 - 1)`.
Prove that sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ.
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A.
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
