Question

The value of $sin\left[2ta{n}^{-1}\right(0.75\left)\right]$ is

(a) 0.75 (b) 1.5 (c) 0.96 (d) sin 1.5

Answer 1

(c) We have $sin\left[2ta{n}^{-1}\right(0.75\left)\right]=sin\left(2ta{n}^{-1}\frac{3}{4}\right)[\because 0.75=\frac{75}{100}=\frac{3}{4}]=sin\left[si{n}^{-1}\frac{2.\frac{3}{4}}{1+\frac{9}{16}}\right]=sin\left[si{n}^{-1}\frac{3/2}{25/16}\right]=sin\left[si{n}^{-1}\left(\frac{48}{50}\right)\right]=sin\left[si{n}^{-1}\left(\frac{24}{25}\right)\right]=\frac{24}{25}=0.96$