How do you find the exact value of $cos ({tan}^{- 1} (- 1))$ $cos(tan^-1(-1))$ ?

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Answer 1 · 2016-12-28T14:19:43

$cos ({tan}^{- 1} (- 1)) = \frac{1}{\sqrt{2}}$ $cos(tan^(-1)(-1))=1/sqrt2$

${tan}^{- 1} (- 1)$ $tan^(-1)(-1)$ means an angle whose tangent ratio is $- 1$ $-1$ .

The range of such an angle is $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$ and in this range, we have $tan (- \frac{π}{4}) = - 1$ $tan(-pi/4)=-1$ .

Hence ${tan}^{- 1} (- 1) = - \frac{π}{4}$ $tan^(-1)(-1)=-pi/4$

and $cos ({tan}^{- 1} (- 1)) = cos (- \frac{π}{4}) = \frac{1}{\sqrt{2}}$ $cos(tan^(-1)(-1))=cos(-pi/4)=1/sqrt2$

How do you find the exact value of cos(tan−1(−1))cos(tan^-1(-1))?