Question

√1+sin theta /√1-sin theta = ??
Pls answer it's really urgent
Ch - Introduction to Trigonometry
Class 10th ​

Answer 1

given:-

$\implies \: \frac{ \sqrt{1 + \sin \theta } }{ \sqrt{1 - \sin \theta } }$

find:-

• given quantity value

★SOLUTION:-

$\implies \: \frac{ \sqrt{1 + \sin \theta } }{ \sqrt{1 - \sin \theta } }$

multiply numerator and denometer

$\implies \red{ \sqrt{1 + \sin \theta} }$

to We get

$\implies \: \frac{ (\sqrt{1 + \sin \theta) }( \sqrt{1 + \sin \theta } ) }{ (\sqrt{1 - \sin \theta })( \sqrt{1 + \sin \theta }) }$

we know that formula of

$\implies \star \pink{(a + b)(a - b) = a {}^{2} - b {}^{2} }$

and

$\implies \star \pink{ \sqrt{a} . \sqrt{a} = a }$

$\implies \: \frac{1 + \sin \theta }{√1 - \sin {}^{2} \theta }$

we know that formula of

$\implies \star\pink{1 - \sin {}^{2} \theta = \cos \theta }$

$\implies \: \frac{1 + \sin \theta }{ \cos {}{} \theta }$

can we be written as

$\implies \: \frac{1}{ \cos \theta } + \frac{ \sin \theta }{ \cos {}^{} \theta }$

we know that

$\implies \star \pink{ \tan \theta = \frac{ \sin \theta }{ \cos \theta} } \\ \\ and \\ \\ \implies \star \pink{ \frac{1}{ \cos( \theta) } = \sec\theta }$

$\implies \red{ \sec {}^{} \theta + \tan \theta }$

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I hope it helps you