Question

Sin³A + cos³A/ sin A + cos A=cosecA*secA-1/cosecAcseA​

Answer 1

We have to prove that, $\dfrac{\sin^3 A+\cos^3 A}{\sin A+\cos A} =\dfrac{\csc A\sec A-1}{\csc A\sec A}.$

L.H.S. = $\dfrac{\sin^3 A+\cos^3 A}{\sin A+\cos A}$

Using the algebraic identity:

$a^{3} +b^{3} =(a+b)(a^2-ab+b^2)$

= $\dfrac{(\sin A+\cos A)(\sin^2 A-\sin A\cos A+\cos^2 A)}{\sin A+\cos A}$

= $\sin^2 A-\sin A\cos A+\cos^2 A$

Using the trigonometric identity:

$\sin^2 A+\cos^2 A=1$

= $1-\sin A\cos A$

= $1-\dfrac{1}{\csc A} \dfrac{1}{\sec A}$

= $\dfrac{\csc A\sec A-1}{\csc A\sec A}$

= L.H.S., proved.

Hence, $\dfrac{\sin^3 A+\cos^3 A}{\sin A+\cos A} =\dfrac{\csc A\sec A-1}{\csc A\sec A},$ proved.