Question

tan-1(1/2)=π\4-1/2 cos-1(4/5)
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Answer 1

Answer:

Let 12cos−1(ab)=x

Then, ab=cos2x→(1)

Now,

L.H.S.=tan(π4+12cos−1(ab))+tan(π4−12cos−1(ab))

=tan(π4+x)+tan(π4−x)

=1+tanx1−tanx+1−tanx1+tanx

=(1+tanx)2+(1−tanx)21−tan2x

=1+tan2x+2tanx+1+tan2x−2tanx1−tan2x

=2(1+tan2x)1−tan2x

We know, tan2x=sec2x−1

So, our expression becomes,

=2sec2x2−sec2x

=2cos2x2−1cos2x

=22cos2x−1

=2cos2x

=2ab (From (1))

=2ba=R.H.S.