question_answer

Prove the identify:  \[(sec\text{ }A-cos\text{ }A)\,.\,(cot\text{ }A+tan\text{ }A)=tan\text{ }A\,.\,\,sec\text{ }A\].

Answer:

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

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

\[=\left( \frac{1-{{\cos }^{2}}A}{\cos A} \right)\left( \frac{{{\cos }^{2}}A+{{\sin }^{2}}A}{\sin A\cos A} \right)\]

\[=\frac{{{\sin }^{2}}A}{\cos A}\times \frac{1}{\sin A\cos A}\]               \[[\because \,\,co{{s}^{2}}A+si{{n}^{2}}A=1]\]

\[=\frac{\sin A}{\cos A}\times \frac{1}{\cos A}\]

\[=\tan A.\sec A=R.H.S.\]                                   Hence Proved.