Question

for which acute angle A cosA/1-sinA+cosA/1+sinA=4true​

Answer 1

Answer:

$sec \: a \: = 2$

$consider \: l.h.s. \frac{cos \: a}{1 - sin \: a} + \frac{cos \: a}{1 + \cos a }$

Taking cos A as common outside , we get

$\cos \: a \: ( \frac{1}{1 - \sin \: a } + \frac{ 1}{1 + \sin \: a} )$

$= \cos \: a \: ( \frac{2}{1 - {sin}^{2} \: a } )$

$= \: cos \: a \: ( \frac{2}{ {cos \: }^{2} } )$

$= \frac{2}{cos \: a} = 2 \: sec \: a$

R.H.S =4

$2sec \: a \: = 4$

Then ,

$\boxed{ \to \: sec \: a \: = \: 2}$

Hope it helps you

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Answer 2

Answer:

Answer:

sec \: a \: = 2seca=2

consider \: l.h.s. \frac{cos \: a}{1 - sin \: a} + \frac{cos \: a}{1 + \cos a }considerl.h.s.

1−sina

cosa

+

1+cosa

cosa

Taking cos A as common outside , we get

\cos \: a \: ( \frac{1}{1 - \sin \: a } + \frac{ 1}{1 + \sin \: a} )cosa(

1−sina

1

+

1+sina

1

)

= \cos \: a \: ( \frac{2}{1 - {sin}^{2} \: a } )=cosa(

1−sin

2

a

2

)

= \: cos \: a \: ( \frac{2}{ {cos \: }^{2} } )=cosa(

cos

2

2

)

= \frac{2}{cos \: a} = 2 \: sec \: a=

cosa

2

=2seca

R.H.S =4

2sec \: a \: = 42seca=4

Then ,

\boxed{ \to \: sec \: a \: = \: 2}

→seca=2