Advertisements
Advertisements
Question
If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`
Advertisements
Solution
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
RELATED QUESTIONS
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
Write the value of tan10° tan 20° tan 70° tan 80° .
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
If tanθ `= 3/4` then find the value of secθ.
\[\frac{1 - \sin \theta}{\cos \theta}\] is equal to
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
Prove that cot2θ × sec2θ = cot2θ + 1
