Question

tan alpha/√1+tan²alpha =​

Answer 1

Step-by-step explanation:

given, tan

−1

(

1+x

2

−

1−x

2

1+x

2

+

1−x

2

)=α

Put, x

2

=cos2α

⇒tan

−1

(

1+cos2α

−

1−cos2α

1+cos2α

+

1−cos2α

)=α

⇒tan

−1

(

2cos

2

α

−

2sin

2

α

2cos

2

α

+

2sin

2

α

)=α

⇒tan

−1

(

cosα−sinα

cosα+sinα

)=α

⇒tan

−1

(

1−tanα

1+tanα

)=α

⇒tan

−1

⎝

⎛

1−tan

4

π

tanα

tan

4

π

+tanα

⎠

⎞

=α

⇒tan

−1

(tan(

4

π

+α))=α

⇒

4

π

+α=α

⇒

4

π

+

2

1

cos

−1

x

2

=α

⇒

2

1

cos

−1

x

2

=α−

4

π

⇒cos

−1

x

2

=2α−

2

π

⇒x

2

=cos(2α−

2

π

)=cos(−(

2

π

−2α))=cos(

2

π

−2α)=sin2α

∴x

2

=sin2α

Answer 2

Answer:

answer for the given problem is given