Let x=sin−1(35)
⇒sinx=35
Now,
cosx=√1−sin2x=√1−925=√1625=45
Thus,
tanx=sinxcosx=34
So, x=tan−134
i.e., sin−1(35)=tan−134
Now, from given equation
L.H.S.=2sin−135
=2tan−1(34)
Using 2tan−1x=tan−1(2x1−x2)
2tan−1(34)=tan−1⎛⎜
⎜
⎜
⎜
⎜⎝2(34)1−(34)2⎞⎟
⎟
⎟
⎟
⎟⎠
=tan−1(247)=R.H.S.
Hence proved.