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Mathematics Question on Inverse Trigonometric Functions

Find the value of tan-1(1)(1)+cos-1(−12)(-\frac{1}{2})+sin-1(−12)(-\frac{1}{2})

Answer

Let tan-1(1)=x.(1)=x.
Then, tan x=1=tan(π4).tan \,x=1=tan\bigg(\frac{\pi}{4}\bigg).
tan-1(1)=π4(1)=\frac{\pi}{4}
Let cos-1(−12)=y,\bigg(-\frac{1}{2}\bigg)=y,
Then cos y=−12=cos(π3)=cos(π−π3)=cos(2π3).cos\,y =-\frac{1}{2}=cos\bigg(\frac{\pi}{3}\bigg)=cos\bigg(\pi-\frac{\pi}{3}\bigg)=cos\bigg(\frac{2\pi}{3}\bigg).
so cos-1(−12)=2π3.\bigg(-\frac{1}{2}\bigg)=\frac{2\pi}{3}.
Let sin-1(−12)=z,\bigg(-\frac{1}{2}\bigg)=z,,
Then sin z=−12=−sin(π6)=sin(−π6).-\frac{1}{2}=-sin\bigg(\frac{\pi}{6}\bigg)=sin\bigg(-\frac{\pi}{6}\bigg).
so sin-1(−12)=−π6\bigg(-\frac{1}{2}\bigg)=-\frac{\pi}{6}

Therefore tan-1(1)(1)+cos-1(−12)(-\frac{1}{2})+sin-1(−12)(-\frac{1}{2})=π4+2π3−π6=3π+8π−2π12=3π4.\frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}=\frac{3\pi+8\pi-2\pi}{12}=\frac{3\pi}{4}.