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
संबंधित प्रश्न
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following trigonometric identities.
`tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`
Prove the following trigonometric identities.
`(1 + cos A)/sin A = sin A/(1 - cos A)`
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
Prove the following identity :
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
Find the value of ( sin2 33° + sin2 57°).
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
