Question

√secx+1/secx-1=1/cosec x -cotx prove the following

Answer 1

To prove.

$\sqrt{ \frac{sec + 1}{sec - 1} } = \frac{1}{cosec} - cot$

$\sqrt{ \frac{sec + 1}{sec - 1} } \times \sqrt{ \frac{sec - 1}{sec - 1} }$

$\sqrt{ \frac{sec {}^{2} - 1}{(sec - 1) {}^{2} } }$

$\sqrt{ \frac{ {tan}^{2} }{ ({sec - 1)}^{2} } }$

$\frac{tan}{sec - 1}$

$\frac{sin}{cos} \div ( \frac{1}{cos} - 1)$

$\frac{sin}{cos} \div \frac{1 - cos}{cos}$

$\frac{sin}{cos} \times \frac{cos}{1 - cos}$

$\frac{sin}{1 - cos}$

$sin - cot$

$\frac{1}{sin} - cot$

$\huge \mathfrak \red{hence \: proved}$

Answer 2

Step-by-step explanation:

this is the answer please mark me as branliest