sinx-sinxxvl +sinx -V1- sinx4

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Solution

We have to prove that cot −1 ( 1+sinx + 1−sinx 1+sinx − 1−sinx )= x 2 , x∈( 0, π 4 ).

Solve left hand side of the given equation by rationalizing the angle.

cot −1 ( 1+sinx + 1−sinx 1+sinx − 1−sinx )= cot −1 ( ( 1+sinx + 1−sinx ) 2 ( 1+sinx ) 2 − ( 1−sinx ) 2 ) = cot −1 ( 1+sinx+1−sinx+2 ( 1+sinx )( 1−sinx ) 1+sinx−1+sinx ) = cot −1 ( 2( 1+ 1− sin 2 x ) 2sinx ) = cot −1 ( 1+cosx sinx )

Further solving,

cot −1 ( 1+sinx + 1−sinx 1+sinx − 1−sinx )= cot −1 ( 2 cos 2 x 2 2sin x 2 cos x 2 ) = cot −1 ( cot x 2 ) = x 2

Hence, it is proved that cot −1 ( 1+sinx + 1−sinx 1+sinx − 1−sinx )= x 2 .

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A

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