Question

please prove the above.​

Answer 1

$\:\:\:\: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \huge\mathcal{\bf{\underline{\underline{\huge\mathcal{Answer}}}}}$

$\large\mathcal\red{solution}$

$L.H.S=\frac{ \cot(a) + \cosec(a) - 1 }{ \cot(a) - \cosec(a) + 1} \\ = > \frac{ \frac{ \cos(a) }{ \sin(a) } + \frac{1}{ \sin(a) } - 1 }{ \frac{ \cos(a) }{ \sin(a) } - \frac{1}{ \sin(a) } + 1 } \\ = > \frac{ \frac{ \cos(a) + 1 - \sin(a) }{ \sin(a) } }{ \frac{ \cos(a) - 1 + \sin(a) }{ \sin(a) } } \\ = > \frac{\cos(a) + 1 - \sin(a) }{\cos(a) - 1 + \sin(a) }\\</p><p>=>\frac{\cos{}^{2}(\frac{a}{2})-\sin{}^{2}(\frac{a}{2})+\cos{}^{2}(\frac{a}{2})+\sin{}^{2}(\frac{a}{2})-\sin(a)}{\cos{}^{2}(\frac{a}{2})-\sin{}^{2}(\frac{a}{2})-\cos{}^{2}(\frac{a}{2})-\sin{}^{2}(\frac{a}{2})+\sin(a)}\\</p><p></p><p>=≥\frac{2\cos{}^{2}(\frac{a}{2})-2\sin(\frac{a}{2})\cos(\frac{a}{2})}{2\sin(\frac{a}{2})\cos(\frac{a}{2})-2\sin{}^{2}(\frac{a}{2})}\\</p><p></p><p>=>\frac{2\cos(\frac{a}{2})(\cos(\frac{a}{2})-\sin(\frac{a}{2})}{2\sin(\frac{a}{2})(\cos(\frac{a}{2})-\sin(\frac{a}{2})}\\</p><p></p><p>=>\frac{\cos(\frac{a}{2})}{\sin(\frac{a}{2})}\\</p><p>=>\cot(\frac{a}{2})\\\\</p><p></p><p>R.H.S=\frac{1+cos(a)}{sin(a)}\\</p><p></p><p>=\frac{2\cos{}^{2}(\frac{a}{2})}{2\sin(\frac{a}{2})\cos(\frac{a}{2})}\\</p><p>=\frac{\cos(\frac{a}{2})}{\sin(\frac{a}{2})}\\</p><p>=\cot(\frac{a}{2})\\</p><p></p><p></p><p>therefore.....L.H.S=R.H.S....(proved)</p><p></p><p></p><p>$

$\large\mathcal\red{hope\: this \: helps \:you......}$