Question

If sec a + tan a = p then the value of sin a is

Answer 1

Answer:

$seca + tana = p$

or, 1/cosa+sina/cosa=p

or, 1+sina/cosa=p

or, 1+sina=pcosa

:.sina=pcosa-1(Answer)

hope you understand :-)

Answer 2

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Given that,

$\sf{sec \theta \: + \: tan \theta \: = p.................(1)}$

Also,

$\sf{sec \theta \: - \: tan \theta \: = \frac{1}{p}..............(2) }$

Adding equations (1) and (2),we get:

$\sf{2sec \theta = p + \frac{1}{p} } \\ \\ \huge{\rightarrow \: \sf{sec \theta = \frac{p {}^{2} + 1}{2p} }}$

We know that,

$\sf{cos \theta \: = \frac{1}{sec \theta} } \\ \\ \huge{\implies \: \sf{cos \theta = \frac{2p}{p {}^{2} + 1} }}$

Using the relation,sin²∅ = 1 - cos²∅

$\sf{sin {}^{2} \theta \: = 1 - ( \frac{2p}{p {}^{2} + 1} ) {}^{2} } \\ \\ \implies \: \sf{sin {}^{2} \theta = \frac{p {}^{4} + 2p {}^{2} + 1 - 4p {}^{2} }{(p {}^{2} + 1) {}^{2} } } \\ \\ \implies \: \sf{sin {}^{2} \theta \: = \frac{(p {}^{2} - 1) {}^{2} }{(p {}^{2} + 1) } } \\ \\ \implies \: \underline{\boxed{ \sf{sin \: \theta = \pm \frac{p {}^{2} - 1 }{p {}^{2} + 1 } - }}}$