Question

Please APPROPRIATE RIGHT ANSWERS ONLY​

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: find : value \: of \: \\ tan \: x + sec \: x \\ \\ given \: that : \\ sin \: x = \frac{p}{q} \\ \\ we \: know \: that \\ cos {}^{2} \: x = 1 - sin {}^{2} \: x \\ = 1 - ( \frac{p}{q} \: ) {}^{2} \\ \\ = 1 - \frac{p {}^{2} }{q {}^{2} } \\ \\ = \frac{q {}^{2} - p {}^{2} }{q {}^{2} } \\ \\ cos \: x = \frac{ \sqrt{q {}^{2} - p {}^{2} } }{q} \\ \\ thus \: then \\ sec \: x = \frac{q}{ \sqrt{q {}^{2} - p {}^{2} } }$

$moreover \: \\ we \: know \: that \\ \\ tan {}^{2} \: x = sec {}^{2} \: x - 1 \\ \\ = \frac{q {}^{2} }{q {}^{2} - p {}^{2} } - 1 \\ \\ = \frac{q {}^{2} - q {}^{2} + p {}^{2} }{q {}^{2} - p {}^{2} } \\ \\ = \frac{p {}^{2} }{q {}^{2} - p {}^{2} } \\ \\ tan \: x = \frac{p}{ \sqrt{q {}^{2} - p {}^{2} } }$

$thus \: then \\ tan \: x + sec \: x = \frac{p}{ \sqrt{q {}^{2} - p {}^{2} } } + \frac{q}{ \sqrt{q {}^{2} - p {}^{2} } } \\ \\ = \frac{p + q}{ \sqrt{q {}^{2} - p {}^{2} } } \\ \\ = \frac{( \sqrt{p + q} \: )( \sqrt{p + q} \: ) }{ \sqrt{(q + p)(q - p)} } \\ \\ = \sqrt{ \frac{(p + q)(p + q)}{(q + p)(q - p)} } \\ \\ = \sqrt{ \frac{p + q}{q - p} } \\ \\ = \sqrt{ \frac{q + p}{q - p} }$