Find the value of tan -1 1 -cos -1 - 1 - 2 -sin -1 - 1 - 2 -

tan^ 1 ( 1 ) +cos^ 1 ( 1 / 2 #160;) +sin^ 1 ( 1 / 2 #160;) #160; =π #160;/ 4 + 2π #160;/ 3 π #160;/ 6 #160; = 3π +8π ...

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Standard XII

Mathematics

Sin(A+B)Sin(A-B)

Find the valu...

Question

Find the value of ${tan}^{- 1} (1) + {cos}^{- 1} (- \frac{1}{2}) + {sin}^{- 1} (- \frac{1}{2}) .$

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Solution

${tan}^{- 1} (1) + {cos}^{- 1} (- \frac{1}{2}) + {sin}^{- 1} (- \frac{1}{2})$
$= \frac{π}{4} + \frac{2 π}{3} - \frac{π}{6}$

$= \frac{3 π + 8 π - 2 π}{12} = \frac{9 π}{12} = \frac{3 π}{4}$

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Q. Find the values of the following :

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Find the value of tan−1(1)+cos−1(−12)+sin−1(−12).

tan−1(1)+cos−1(−12)+sin−1(−12)=π4+2π3−π6=3π+8π−2π12=9π12=3π4

Find the value of ${tan}^{- 1} (1) + {cos}^{- 1} (- \frac{1}{2}) + {sin}^{- 1} (- \frac{1}{2}) .$

${tan}^{- 1} (1) + {cos}^{- 1} (- \frac{1}{2}) + {sin}^{- 1} (- \frac{1}{2})$
$= \frac{π}{4} + \frac{2 π}{3} - \frac{π}{6}$

$= \frac{3 π + 8 π - 2 π}{12} = \frac{9 π}{12} = \frac{3 π}{4}$