Question

if secø+tanø=p, then what is the value of secø-tanø? ​

Answer 1

Answer:

$\sec( \alpha ) - \tan( \alpha ) = \frac{1}{p}$

Step-by-step explanation:

$p = \frac{ \sec( \alpha ) + \tan( \alpha ) }{1} \\ \\ \frac{1}{p} = \frac{1}{ \sec( \alpha ) + \tan( \alpha ) } \\ \\ \frac{1}{p} = \frac{1}{ \sec( \alpha ) + \tan( \alpha ) } \times \frac{ \sec( \alpha ) - \tan( \alpha ) }{sec( \alpha ) - \tan( \alpha )} \\ \\ \frac{1}{p} = \frac{ \sec( \alpha ) - \tan( \alpha ) }{ { \sec( \alpha ) }^{2} - { \tan( \alpha ) }^{2} } \\ \\ \frac{1}{p} = \sec( \alpha ) - \tan( \alpha ) \\ \\ \\ therefore \: \: \: \: \: \sec( \alpha ) - \tan( \alpha ) = \frac{1}{p}$