Question

$\frac{{(1 + sin)}^{2} + {(1 - sin)}^{2} }{ {2cos}^{2} } = \frac{1 + {sin}^{2} }{1 - {sin}^{2} }$
​

Answer 1

Answer:

We know that,

(

1

)

1

+

tan

2

θ

=

sec

2

θ

(

2

)

1

−

sin

2

θ

=

cos

2

θ

(

3

)

cos

θ

=

1

sec

θ

Here,

√

1

+

tan

2

x

√

1

−

sin

2

x

=

sec

2

x

Let,

L

H

S

=

√

1

+

tan

2

x

√

1

−

sin

2

x

Using

(

1

)

and

(

2

)

we get

L

H

S

=

√

sec

2

x

√

cos

2

x

L

H

S

=

sec

x

cos

x

L

H

S

=

sec

x

(

1

sec

x

)

L

H

S

=

sec

x

⋅

sec

x

L

H

S

=

sec

2

x

∴

L

H

S

=

R

H

S

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