Question

√1+sinA/1-sinA = secA + tan A
​

Answer 1

Answer:

Answer

LHS=

1−sinA

1+sinA

=

1−sinA

1+sinA

×

1+sinA

1+sinA

=

(1+sinA)(1−sinA)

(1+sinA)

2

=

1−sin

2

A

(1+sinA)

2

=

cos

2

A

(1+sinA)

2

=

(

cosA

1+sinA

)

2

=

cosA

1+sinA

=

cosA

1

+

cosA

sinA

=secA+tanA

=RHS

Answer 2

★verified:-

$\implies \: \red{\sqrt{ \frac{1 + \sin A}{1 - \sin A } } } = \orange{ \sec \: A - \tan \: A }$

★solution:-

$\implies \blue{lhs}$

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

multiplie (1+sin A) numerator and denometer

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

we know that

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

can be written as

$\implies \: \frac{ \sqrt{(1 + \sin \: A) {}^{2} } }{ \sqrt{1 - \sin {}^{2} A } }$

we know that

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

$\implies \: \frac{1 + \sin A}{ \sqrt{ \cos {}^{2}A } }$

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

can we be written as

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

we know that

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

$\implies \orange{ \sec A+ \tan A}(hence \: proved )$

lhs=rhs

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I hope it helps you