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Question Number 189375 by moh777 last updated on 15/Mar/23

s h o w t h a t :

\frac{1}{c s c x + c o t x} = c s c x - c o t x

Answered by Ar Brandon last updated on 15/Mar/23

\frac{1}{\csc x + \cot x} = \frac{\sin x}{1 + \cos x} = \frac{\sin x (1 - \cos x)}{1 - \cos^{2} x}

= \frac{\sin x (1 - \cos x)}{\sin^{2} x} = \frac{1 - \cos x}{\sin x}

= \csc x - \cot x

Answered by BaliramKumar last updated on 15/Mar/23

L H S = \frac{1}{c o s e c x + c o t x} = \frac{{c o s e c}^{2} x - {c o t}^{2} x}{c o s e c x + c o t x}

= \frac{(c o s e c x + c o t x) (c o s e c x - c o t x)}{(c o s e c x + c o t x)}

= c o s e c x - c o t x = R H S

Answered by manxsol last updated on 15/Mar/23

r e c a l l {c o s e c}^{2} = 1 + {c o t}^{2} x

\frac{1}{c s c x + c o t x} \times \frac{c s c x - c o t x}{c s c x - c o t x} =

\frac{c s c x - c o t x}{{c s c}^{2} x - {c t}^{2} x} = c s c x - c o t x

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