Question

$\frac{ \tan(a) }{1 - \cot(a) } + \frac{ \cot(a) }{1 - \ \tan(a) } = 1 + \sec(a) \csc(a)$
prove this identity. ​

Answer 1

$\sf\underline\pink{Question}$

 We are given to prove the following trigonometric equality :

$\tt\sf\dfrac{\tan A}{1\cot A}+\dfrac{\cot A}{1-\tan A}=1+\sec A\cos A.$

$\tt\sf\underline\purple{Using\:Trigonometry \:Formulas}$

$\sf\tt(i)\tan\theta=\dfrac{\sin\theta}{\cos\theta},\\\\\\(ii)\sf\tt\sin^2\theta+\cos^2\theta=1.$

$\sf\underline\red{Proof}$

$L.H.S.\\\\\\\tt\sf\dfrac{\tan A}{1-\cot A}+\sf\dfrac{\cot A}{1-\tan A}\\\\\\=\tt\dfrac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}+\dfrac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}\\\\\\=\tt\sf\dfrac{\sin A}{\cos A}\times\dfrac{\sin A}{\sin A-\cos A}+\dfrac{\cos A}{\sin A}\times\dfrac{\cos A}{\cos A-\sin A}\\\\\\=\tt\sf\dfrac{\sin^2 A}{\cos A(\sin A-\cos A)}-\dfrac{\cos^2 A}{\sin A(\sin A-\cos A)}\\\\\\=\tt\sf\dfrac{\sin^3 A-\cos^3 A}{\sin A\cos A(\sin A-\cos A)}\\\\\\=\tt\sf\dfrac{(\sin A-\cos A)(\sin^2 A+\sin A\cos A+\cos^2 A)}{\sin A\cos A(\sin A-\cos A)}\\\\\\=\tt\sf\dfrac{\sin A\cos A+1}{\sin A\cos A}\\\\\\=1+\sec A\cos A\\\\=R.H.S.$

HENCE PROVED!!