Question #37594

Prove it plz for me?
1+sinθ/1-sinθ = cosecθ+1/cosecθ-1
plz solve it in detail!

Expert's answer

Answer on Question#37594 - Math - Trigonometry


1+sinθ1sinθ=cosecθ+1cosecθ11 + \frac{\sin \theta}{1 - \sin \theta} = \cosec \theta + \frac{1}{\cosec \theta - 1}


Solution.


cosecθ=1sinθ.\cosec \theta = \frac{1}{\sin \theta}.


We obtain


1+sinθ1sinθ=1sinθ+11sinθ1;1 + \frac{\sin \theta}{1 - \sin \theta} = \frac{1}{\sin \theta} + \frac{1}{\frac{1}{\sin \theta} - 1};1+sinθ1sinθ=1sinθ+sinθ1sinθ;1 + \frac{\sin \theta}{1 - \sin \theta} = \frac{1}{\sin \theta} + \frac{\sin \theta}{1 - \sin \theta};1=1sinθ1 = \frac{1}{\sin \theta}sinθ=1\sin \theta = 1 \rightarrowθ=π2+2πn, where nZ.\theta = \frac{\pi}{2} + 2\pi n, \text{ where } n \in \mathbb{Z}.

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