Prove that: ((cot A + tan A - 1)(sin A + cos A)) / (sin³ A + cos³ A) = sec A × cosec A

Left-hand side:

[(cot A + tan A − 1)(sin A + cos A)] ------------------------------------------------ sin³ A + cos³ A

Write cot A and tan A in terms of sine and cosine:

= [(cos A/sin A + sin A/cos A − 1)(sin A + cos A)] ---------------------------------------------------------------- sin³ A + cos³ A

Combine the first bracket:

cos A/sin A + sin A/cos A = (cos² A + sin² A)/(sin A cos A)

Using: sin² A + cos² A = 1

= 1/(sin A cos A)

Therefore:

cot A + tan A − 1 = 1/(sin A cos A) − 1

= [1 − sin A cos A]/(sin A cos A)

Also:

sin³ A + cos³ A = (sin A + cos A)(sin² A − sin A cos A + cos² A)

Using: sin² A + cos² A = 1

sin³ A + cos³ A = (sin A + cos A)(1 − sin A cos A)

Substitute these results:

Left-hand side

= {[(1 − sin A cos A)/(sin A cos A)](sin A + cos A)} ---------------------------------------------------------------- {(sin A + cos A)(1 − sin A cos A)}

Cancel the common factors:

= 1/(sin A cos A)

= (1/cos A)(1/sin A)

= sec A × cosec A

= Right-hand side

Hence proved.

Every algebraic and trigonometric step is shown on a separate line. The identities used are sin²A + cos²A = 1 and a³+b³ = (a+b)(a²−ab+b²).