Question

Proof the following ​

Answer 1

Step-by-step explanation:

$\sqrt{ \frac{sec \: x + 1}{sec \: x - 1} } = \frac{1}{cosec \: x - cot \: x} \\ LHS = \sqrt{ \frac{sec \: x + 1}{sec \: x - 1} } \\ multiplying \: numerator \: and \: \\ denominator \: by \: \sqrt{sec \: x - 1} \: we \: find : \\ = \sqrt{ \frac{sec \: x + 1}{sec \: x - 1} } \times \sqrt{ \frac{sec \: x - 1}{sec \: x - 1} } \\ = \sqrt{ \frac{sec^{2} \: x - {1}^{2} }{(sec \: x - 1)^{2} } } \\ = \sqrt{ \frac{sec^{2} \: x - {1} }{(sec \: x - 1)^{2} } } \\ = \sqrt{ \frac{tan^{2} \: x }{(sec \: x - 1)^{2} } } \\ = \frac{tan \: x }{sec \: x - 1} \\ = \frac{ \frac{sin \: x}{cos \: x} }{ \frac{1}{cos \: x} - 1 } \\ = \frac{ \frac{sin \: x}{cos \: x} }{ \frac{1 - cos \: x}{cos \: x} } \\ = \frac{ \frac{sin \: x}{cos \: x} \times cos \: x}{ 1 - cos \: x } \\ = \frac{ sin\: x \: }{ 1 - cos \: x } \\ = \frac{ \frac{sin\: x}{sin\: x}}{ \frac{1 - cos \: x}{sin\: x} }.. (Dividing \:Nr. \:\& \:Dr. \:by \:sin\: x) \\ = \frac{1}{ \frac{1}{sin\: x} - \frac{cos\: x}{sin\: x} } \\ = \frac{1}{cosec \: x - cot \: x} \\ = RHS$

Thus Proved