question_answer

Prove that: \[(1+cot\text{ }A+tan\text{ }A).\text{(}sin\text{ }A-cos\text{ }A)\]\[=\frac{{{\sec }^{3}}A-\cos e{{c}^{3}}A}{{{\sec }^{2}}A.\cos e{{c}^{2}}A}\]

Answer:

\[L.H.S.=(1+cot\text{ }A+tan\text{ }A)\text{ (}\sin \text{ }A-cos\text{ }A)\]

\[=\left( 1+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A} \right)(sin\,A-cos\,A)\]

\[=\left( \frac{\sin \,A\cos A+{{\cos }^{2}}A+{{\sin }^{2}}A}{\sin A.\cos A} \right)(sinA-cosA)\]

\[=\frac{{{\sin }^{3}}A-{{\cos }^{3}}A}{\sin A.\cos A}\]

\[[\text{Using}\,{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})\]

\[=\frac{\frac{{{\sin }^{3}}A}{{{\sin }^{3}}A.{{\cos }^{3}}A}-\frac{{{\cos }^{3}}A}{{{\sin }^{3}}A.{{\cos }^{3}}A}}{\frac{\sin A\cos A}{{{\sin }^{3}}A.{{\cos }^{3}}A}}\]

[Dividing Num. & Denom. by \[si{{n}^{3}}A.\,co{{s}^{3}}A\] ]

\[=\frac{{{\sec }^{3}}A-\cos e{{c}^{3}}A}{{{\sec }^{2}}A.\cos e{{c}^{2}}A}=R.H.S.\]                   Hence Proved.