Inverse Trigonometric Functions

Given that,
cot−1(√cos‌x)−tan−1(cos‌x)=x.........(1)
We know,
cot−1x+tan−1x=‌

π

2

∴cot−1(√cos‌x)+tan−1(√cos‌x)=‌

π

2

.............(2)
Adding (1) and (2), we get,
2cot−1(√cos‌x)=x+‌

π

2

⇒cot−1(√cos‌x)=‌

x

2

+‌

π

4

⇒√cos‌x=cot(‌

n

2

+‌

π

4

)
⇒√cos‌x=‌

cot‌ x 2 −1

1+cot‌ x 2

⇒√cos‌x=‌

cos‌ x 2 −sin‌ x 2

cos‌ x 2 +sin‌‌ x 2

Squaring both sides we get,
⇒cos‌x=‌

1−2sin‌ x 2 ‌cos‌ x 2

1+2sin‌‌ x 2 ‌cos‌ x 2

⇒cos‌x=‌

1−sin‌x

1+sin‌x

⇒‌

1−tan2 z 2

1+tan2‌ z 2

=‌

1−sin‌x

1+sin‌x

Applying compounds and dividendo rule,
⇒‌

2sin‌x

2

=‌

2tan2 x 2

2

⇒sin‌x=tan2‌

x

2