Show that: `cos(2tan^(-1)1/7)=sin(4^(-1)1/3)`

Answer

N/a

Explanation

We have, `cos(2tan^(-1)'1/7) = sin(4tan^(-1)'1/3)`
`rArr cos[cos^(-1)((1-(1/7)^(2))/(1+(1/7)^(2)))] = sin[2.2tan^(-1)'1/3] , [:' 2tan^(-1)x = cos^(-1)((1-x^(2))/(1+x^(2)))]`
`rArrcos[cos^(-1)'(((48)/(49))/((50)/(49)))]=2sin[2.(tan^(-1)'(2/3)/(1-(1/3)^(2)))] , [:'2tan^(-1)x=tan^(-1)((2x)/(1-x^(2)))]`
`rArr cos[cos^(-1)'((48xx49)/(50xx49))]=sin[2tan^(-1)(18/24)]` ltb rgt `rArrcos[cos^(-1)(4/25)]=sin(2tan^(-1)'3/4)`
`rArr cos[cos^(-2)(24/25)] sin(sin^(-1)'(2xx3/4)/(1+9/16)),[:' 2tan^(-1)x = sin^(-1)'(2x)/(1+x^(2))]`
`rArr 24/25=sin(sin^(-1)'(3//2)/(25//16))`
` = 24/25=48/50rArr24/25=24/25`
`:. LHS = RHS`