tan(1/2(cos^-1(square root 5/3))
Tan(½cos-1(√5/3))
Let (½cos-1(√5/3)=x
cos-1(√5/3)=2x
(√5/3)=cos 2x
(√5/3)=1-tan2x/1+tan2x
√5 +√5tan2x= 3- 3tan2x
√5tan2x + 3tan2x= 3 - √5
= 3 + √5(tan2x)= 3 - √5
Tan2x = 3 - √5/ 3 + √5
On rationalising we get
Tan2x =( 3 - √5)2/9-5
Tan2x =( 3 - √5)2/4
Tan x =( 3 - √5)/2
x =tan-1( 3 - √5)/2
now Tan(½cos-1(√5/3))
=tan x
=tan(tan-1( 3 - √5)/2)
= ( 3 - √5)/2=ans
Hope u liked it J
Let (½cos-1(√5/3)=x
cos-1(√5/3)=2x
(√5/3)=cos 2x
(√5/3)=1-tan2x/1+tan2x
√5 +√5tan2x= 3- 3tan2x
√5tan2x + 3tan2x= 3 - √5
= 3 + √5(tan2x)= 3 - √5
Tan2x = 3 - √5/ 3 + √5
On rationalising we get
Tan2x =( 3 - √5)2/9-5
Tan2x =( 3 - √5)2/4
Tan x =( 3 - √5)/2
x =tan-1( 3 - √5)/2
now Tan(½cos-1(√5/3))
=tan x
=tan(tan-1( 3 - √5)/2)
= ( 3 - √5)/2=ans
Hope u liked it J
