Question

. Prove that (cosec A-cot A)^2=
(1-сos A)/
(1 + cos A)
​

Answer 1

Hey friend,

Here is the solution :-

$( \csc(a) - \cot(a) )^{2} = \frac{1 - \cos(a) }{1 + \cot(a) } \\ \\ \csc(a) - \cot(a) = \sqrt{ \frac{1 - \cos(a) }{1 + \cos(a) } } \\ \\ r.h.s. = \sqrt{ \frac{1 - \cos(a) }{1 \cos(a)} \times \frac{1 - \cos(a) }{1 - \cos(a) } } \: \: \: \: (by \: rationalizing) \\ \\ = > \sqrt{ \frac{(1 - \cos(a)) \: (1 - \cos(a)) }{(1 + \cos(a)) \: (1 - \cos(a)) } } \\ \\ = > \sqrt{ \frac{(1 - \cos(a))^{2} }{ ({1})^{2} -( \cos(a))^{2} } } \\ \\ = > \sqrt{ \frac{(1 - \cos(a)) ^{2} }{1 - \cos^{2}a } } \\ \\ = > \frac{1 - \cos(a) }{1 - \cos^{2} a } \\ \\ \frac{1 - \cos(a) }{ \sin(a) } \: \: \: \: \: \: \: \: \: \: (1 - \cos(a) = \sin(a) ) \\ \\ = > \frac{1}{ \sin(a) } - \frac{ \cos(a) }{ \sin(a) } \\ \\ = > \csc(a) - \cot(a) \: \: \: \: \: (hence \: l.h.s. = r.h.s.)$

✨I hope this will help you....✨