Math, asked by athulyasanjai, 1 year ago

sin^3 A+cos^3 A/(sinA+cosA)+sin^3 A-cos^3 A/(sin A-cosA)

Answers

Answered by Swarup1998
4

Solution :

Now, \frac{sin^{3}A+cos^{3}A}{sinA+cosA}+\frac{sin^{3}A-cos^{3}A}{sinA-cosA}

= \frac{(sinA+cosA)(sin^{2}A+cos^{2}A-sinA\:cosA)}{(sinA+cosA)}

+ \frac{(sinA-cosA)(sin^{2}A+cos^{2}A+sinA\:cosA)}{(sinA-cosA)}

= (1 - sinA cosA) + (1 + sinA cosA)

= 1 - sinA cosA + 1 + sinA cosA

= 2 (Ans.)

Rules :

  • sin²A + cos²A = 1
  • sinⁿA = (sinA)ⁿ
Answered by nalinsingh
3

Answer:

Step-by-step explanation:

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