Prove the given identity, where the angles involved are acute ang… - Vidyadip

Question

Prove the given identity, where the angles involved are acute angles for which the expressions are defined.$ \frac{{\tan A }}{{1 - \cot A }} + \frac{{\cot A }}{{1 - \tan A }}$ = 1 + sec A cos ecA
[Hint: Write the expression in terms of sin $\theta$ and cos $\theta$]

✓

Answer

LHS-$\frac{{\tan A}}{{1 - \cot A }} + \frac{{\cot A }}{{1 - \tan A }}$
=$\frac{{\tan A}}{{1 - \frac{1}{{\tan A}}}} + \frac{{\frac{1}{{\tan A}}}}{{1 - \tan A}}$
=$\frac{{\tan A}}{{\frac{{\tan A - 1}}{{\tan A}}}} + \frac{1}{{\tan A(1 - \tan A)}}$
=$\frac{{{{\tan }^2}A}}{{\tan A - 1}} + \frac{1}{{\tan A(1 - \tan A)}}$
=$\frac{{{{\tan }^3}A - 1}}{{\tan A(\tan A - 1)}}$
=$\frac{{(\tan A - 1)({{\tan }^2}A + \tan A + 1)}}{{\tan A(\tan A - 1)}}$ $[a^3-b^3=(a-b)(a^2+ab+b^2)]$
=$\frac{{{{\tan }^2}A + \tan A + 1}}{{\tan A}}$
=$\tan A + 1 + \cot A$
=$\frac{{\sin A}}{{\cos A}} + \frac{{\cos A}}{{\sin A}} + 1$
=$\frac{{{{\sin }^2}A + {{\cos }^2}A}}{{\sin A\cos A}} + 1$
=$\frac{1}{{\sin A\cos A}} + 1$
= sec A cosec A+ 1
= R.H.S

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