Question

soLution
(i) Let ( cos ^{-1} frac{3}{5}=theta, ) where ( theta in[0, pi] )
Then, ( cos theta=frac{3}{5} )
since ( theta in[0, pi], ) we have ( sin theta>0 )
( therefore sin theta=sqrt{1-cos ^{2} theta}=sqrt{1-frac{9}{25}}=sqrt{frac{16}{25}}=frac{4}{5} )
Hence, ( sin left(cos ^{-1} frac{3}{5}right)=sin theta=frac{4}{5} )
(ii) Let ( tan ^{-1} frac{3}{4}=theta, ) where ( theta inleft(frac{-pi}{2}, frac{pi}{2}right) )
Then, ( tan theta=frac{3}{4} )
since ( theta inleft(frac{-pi}{2}, frac{pi}{2}right), ) so ( cos theta>0 )
( cos theta=frac{1}{sec theta}=frac{1}{sqrt{1+tan ^{2} theta}}=frac{1}{sqrt{1+frac{9}{16}}}=sqrt{frac{16}{25}}=frac{4}{5} )
Hence, ( cos left(tan ^{-1} frac{3}{4}right)=cos theta=frac{4}{5} )

# Evaluate: (1) sin (cos-13) (i) cos( tan* ) ce

Solution