Question

prove that (Sin^3A+cos^3A/sinA+cosA) + (sin^3A-cos^3A/sinA-cosA)=2

Answer 1

First break the cube then divide by the denominator and then some terms will be cancelled and will be left with
1+1 = 2 = RHS (proved)

Answer 2

Trigonometry,

We have to prove that
(sin³A+cos³A/sinA+cosA) + (sin³A-cos³A/sinA-cosA)=2

LHS = (sin³A+cos³A/sinA+cosA) + (sin³A-cos³A/sinA-cosA)

= {(sinA+cosA)(sin²A+cos²A-sinA.cosA)}/(sinA+cosA) + {(sinA-cosA)(sin²A+cos²A+sinA.cosA)}/(sinA-cosA)

= (sin²A+cos²A-sinA.cosA) + (sin²A+cos²A+sinA.cosA)

= 1-sinA.cosA+1+sinA.cosA

= 1+1

= 2 = RHS [proved]

That's it
Hope it helped (^_^メ)