Advertisements
Advertisements
Question
Prove the following identity :
`sin^2Acos^2B - cos^2Asin^2B = sin^2A - sin^2B`
Advertisements
Solution
LHS = `sin^2A(1 - sin^2B) - (1 - sin^2A)sin^2B`
= `sin^2A - sin^2A.sin^2B - sin^2B + sin^2A.sin^2B`
= `sin^2A - sin^2B` = RHS
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
Prove the following trigonometric identities.
`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
Show that : tan 10° tan 15° tan 75° tan 80° = 1
Prove the following identities:
`sinA/(1 - cosA) - cotA = cosecA`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
` (sin theta - cos theta) / ( sin theta + cos theta ) + ( sin theta + cos theta ) / ( sin theta - cos theta ) = 2/ ((2 sin^2 theta -1))`
If cos A + cos2 A = 1, then sin2 A + sin4 A =
(1 – cos2 A) is equal to ______.
