Find the value of `sin(cos^(-1)(3/4)-tan^(-1(1/4)))`?

Let `cos^(-1)(3/4)=alpha`, then `cosalpha=3/4`

and `sinalpha=sqrt(1-cos^2alpha)`

= `sqrt(1-9/16)=sqrt7/4`

let `tan^(-1)(1/4)=beta` then we have `tanbeta=1/4`

and `cosbeta=1/secbeta=1/sqrt(1+tan^2beta)=1/sqrt(1+1/16)=4/sqrt17`

and `sinbeta=tanbetaxxcosbeta=1/4xx4/sqrt17=1/sqrt17`

Hence `sin(cos^(-1)(3/4)-tan^(-1)(1/4))`

= `sin(alpha-beta)=sinalphacosbeta-cosalphasinbeta`

= `sqrt7/4xx4/sqrt17-3/4xx1/sqrt17`

= `(4sqrt7-3)/(4sqrt17)`

The given expression
`sin (cos^-1(3/4) - tan^-1(1/4))`
looks like `sin (A-B)`
We also know that
`sin(A-B)=sinAcosB-cosAsinB` ....(1)
where
`A=cos^-1(3/4)` .....(2)
and
`B= tan^-1(1/4)` ......(3)

To calculate `sin A and cos A` we make use of (2)
From (2)
`cos A=cos [cos^-1(3/4)]`
`cos A=3/4` .....(4)
Using the identity
`sin^2A+cos^2A=1`
`sinA=sqrt(1-cos^2A)`
`=>sinA=sqrt(1-(3/4)^2)`
`=>sinA=sqrt(1-9/16)`
`=>sinA=sqrt7/4` ......(5)

To calculate `sin B and cos B` we make use of (3)
From (3)
`tanB= tan[tan^-1(1/4)]`
`=>tanB= 1/4` .....(6)
we know that
`cosB=1/(secB)`
`=>cosB=1/sqrt(1+tan^2B)`
Inserting value of `tanB` we get
`=>cosB=1/sqrt(1+(1/4)^2)`
`=>cosB=1/sqrt(1+1/16)`
`=>cosB=4/sqrt(17)` ......(7)

We know that
`sinB=tanBxxcosB`
Using (6) and (7) we get
`sinB=1/4xx4/sqrt17`
`sinB=1/sqrt17` .....(8)

Inserting calculated values in (1) we get
`sin(A-B)=sqrt7/4xx4/sqrt17-3/4xx1/sqrt17`
`sin(A-B)=(4sqrt7-3)/(4sqrt17)`
Rationalizing the denominator we get
`sin(A-B)=(sqrt17(4sqrt7-3))/(68)`

.-.-.-.-.-.-.-.-.-.-

Alternatively after (4)
In a right angled triangle
`cosA=3/4-="base"/"hypotenuse"`
and
`"perpendicular"=sqrt("hypotenuse"^2-"base"^2)`
`:. "perpendicular"=sqrt(4^2-3^2)`
`=> "perpendicular"=sqrt(7)`
This gives us
`sinA-="perpendicular"/"hypotenuse"=sqrt7/4`
This is same as (5)

Similar steps can be followed after (6) to calculate `sinB and cos B` that `tanB=1/4-="perpendicular"/"base"`
and
`"hypotenuse"=sqrt("perpendicular"^2+"base"^2)`