Question

Please answer this. (correctly) Prove the following: (sec^(6)A+tan^(6)A)/(sec^(2)A+tan^(2)A) = 1+sec^(2)A*tan^(2)A

Answer 1

Answer:

Step-by-step explanation:

Sec⁶A + Tan⁶A / Sec²A + Tan²A

= (Sec²A)³ + (Tan²A)³ / Sec²A + Tan²A

The Numerator is of form a³ + b³, i.e. a³ + b³ = (a+b) (a² - ab + b²)

= (Sec²A + Tan²A)( Sec⁴A + Tan⁴A - Sec²ATan²A) / Sec²A + Tan²A

= Sec⁴A + Tan⁴A - Sec²ATan²A

= Sec²A (Sec²A - Tan²A) + Tan⁴A  

= Sec²A + Tan⁴A         (∵Sec²A - Tan²A = 1)

= 1 + Tan²A + Tan⁴A    (∵Sec²A = 1 +  Tan²A)

= 1 + Tan²A ( Tan²A + 1)

= 1 + Tan²A Sec²A.  

= R.H.S

Hence proved.