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)#
