Question

sin^3ø+cos^3ø÷sinø+cosø=

Answer 1

Correction in question :

( sin³∅ + cos³∅ ) ÷ ( sin∅ + cos∅ ) = ?

Solution :

= ( sin³∅ + cos³∅ ) ÷ ( sin∅ + cos∅ )

Using algebric identity,

⇒ ( a³ + b³ ) = ( a + b ) ( a² + b² - ab )

= [ ( sin∅ + cos∅ ) ( sin²∅ + cos²∅ - sin∅ cos∅ ) ] ÷ ( sin∅ + cos∅ )

= ( sin²∅ + cos²∅ - sin∅ cos∅ )

Using Trigonometric identity,

⇒ ( sin²∅ + cos²∅ ) = 1

= ( 1 - sin∅ cos∅ )

The required answer is ( 1 - sin∅ cos∅ ).

Answer 2

Given :- Sin^3Ø + Cos^3Ø/SinØ+CosØ

Here we will apply algebraic identities :-

=> a^3 + b^3 = (a+b) (a^2 + b^2 - ab)

and Trigonometrical Identities :-

=> Sin^2Ø + Cos^2Ø = 1

= Sin^3Ø + Cos^3Ø/SinØ + CosØ

= (SinØ + CosØ) (Sin^2Ø + Cos^2Ø - SinØ.CosØ)/SinØ + CosØ

= (Sin^2Ø + Cos^2Ø - SinØ. CosØ)

= ( 1 - SinØ.CosØ)

Hence, Sin^3Ø + Cos^3Ø/SinØ+CosØ = 1- SinØ.CosØ