Math, asked by johribhakti, 6 months ago

if sin theeta = a/b than pdove that ( sac theeta + tan theeta) =√b+a/√b-a

Answers

Answered by Ayodhyavasi

Step-by-step explanation:

$let \: here \: theeta = \alpha \\ because \: symbol \: theeta \: is \: not \: available \: in \: keyboard$

then

$\sin( \alpha ) = \frac{a}{b}$

From formula

$sin^{2} \alpha + cos^{2} \alpha = 1$

we get

$cos \alpha = \sqrt{1 - sin^{2} \alpha }$

$cos \alpha = \sqrt{1 - \frac{ {a}^{2} }{ {b}^{2} } }$

$cos \alpha = \sqrt{ \frac{ {b}^{2} - {a}^{2} }{ {a}^{2} } }$

$cos \alpha = \frac{ \sqrt{ {b}^{2} - {a}^{2} } }{a}$

then value of

$sec \alpha + tan \alpha = \frac{1}{cos \alpha } + \frac{sin \alpha }{cos \alpha }$

$\: \: \: \: \: = \frac{1 + sin \alpha }{ \cos( \alpha ) }$

$\: \: \: \: \: \: = \frac{1 + \frac{a}{b} }{ \frac{ \sqrt{ {a}^{2} - {b}^{2} } }{a} }$

$\: \: \: \: \: \: = \frac{ \frac{b + a}{a} }{ \frac{ \sqrt{ {b}^{2} - {a}^{2} } }{a} }$

$\: \: \: \: = \frac{b + a}{ \sqrt{ {b}^{2} - {a}^{2} } }$

$\: \: \: \: \: = \sqrt{ \frac{ {(b + a)}^{2} }{(b + a)(b - a)} }$

$= \: \: \: \: \sqrt{ \frac{b + a}{b - a} }$

$\: \: \: \: \: = \frac{ \sqrt{b + a} }{ \sqrt{b - a} }$

Proved.

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if sin theeta = a/b than pdove that ( sac theeta + tan theeta) =√b+a/√b-a​

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if sin theeta = a/b than pdove that ( sac theeta + tan theeta) =√b+a/√b-a