Question

cos 3A+sin3A/cos A-sin A = 1+2sin2A​

Answer 1

Answer:

Step-by-step explanation:

cos3A = 4cosA - 3cos³A,

sin3A = 3sinA - 4sin³A,

substituting we get, 

(4cos³A - 3cosA + 3sinA - 4sin³A)/cosA - sinA 

= 4(cos³A - sin³A) - 3(cosA-sinA) / cosA - sinA

Since, cos³A - sin³A is of form a³ - b³ => a³ - b³ = (a - b)(a²+ b² + ab)

= 4(cosA - sinA)(cos²A + sin²A + cosAsinA) - 3(cosA-sinA) / cosA - sinA

//Cancel out cosA - SinA from numerator and denominator

= 4(cos²A+sin²A+sinAcosA) - 3

= 4 ( 1 + sinAcosA) - 3  (∵ Sin²A + Cos²A = 1)

= 4 + 4sinAcosA - 3

= 1 + 4sinAcosA

= 1 + 2sin2A.    ( ∵ Sin2A = 2SinACosA)

= R.H.S

Hence proved