Question

if sec0+tan0 = x prove that sin0 = x2-1/x2+1​

Answer 1

Answer:

hope it helps you see the attachment for further information

Answer 2

Step-by-step explanation:

$sec0 + tan0 = x - - - - - - 1\\ \frac{1}{sec0 + tan0} = \frac{1}{x}$

$\frac{sec0 - tan0}{ {sec}^{2}0 - {tan}^{2} 0 } = \frac{1}{x}$

$\sec(0) - \tan(0) = \frac{1}{x}$

since

${ \sec}^{2} 0 - {tan}^{2} 0 = 1$

2sec0 = x+1/x

2tan0 = x-1/x

$\frac{ \tan(0) }{ \sec(0) } = \frac{x + \frac{1}{x} }{x - \frac{1}{x} }$

$\sin(0) = \frac{ {x}^{2} - 1 }{ {x }^{2} + 1 }$