Find the value of tan−1(−31)+cot−1(−31)+tan−1[sin(−2−π)].

tan−1(−31)+cot−1(−31)+tan−1[sin(−2−π)].=tan−1(tan65π)+cot−1(cot3π)+tan−1(−1)=tan−1[tan(π−6π)]+cot−1[cot(3π)]+tan−1[tan(π−4π)]=tan−1(−tan6π)+cot−1(cot3π)+tan−1(−tan4π)⎣⎡∵tan−1(tanx ) =x,x∈(−2π ,2π ),cot−1(cotx ) =x,x∈(0,π ) andtan−1(−x)=−tan−1x ⎦⎤=−6π+3π−4π=12−2π+4π−3π=12−5π+4π=12−π

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Find the value of ^{-1}(- {1}{ 3})+^{-1}(- {1}{ 3})+^{-1}[ (-

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Find the value of $tan^{- 1} (- \frac{1}{3}) + cot^{- 1} (- \frac{1}{3}) + tan^{- 1} [sin (- \frac{- π}{2})]$ .

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tan^{- 1} (- \frac{1}{3}) + cot^{- 1} (- \frac{1}{3}) + tan^{- 1} [sin (- \frac{- π}{2})]

= tan^{- 1} (tan \frac{5 π}{6}) + cot^{- 1} (cot \frac{π}{3}) + tan^{- 1} (- 1)

= tan^{- 1} [tan (π - \frac{π}{6})] + cot^{- 1} [cot (\frac{π}{3})] + tan^{- 1} [tan (π - \frac{π}{4})]

= tan^{- 1} (- tan \frac{π}{6}) + cot^{- 1} (cot \frac{π}{3}) + tan^{- 1} (- tan \frac{π}{4})

⎣ ⎡ ∵ tan^{- 1} (tan x) = x, x \in (- \frac{π}{2}, \frac{π}{2}), cot^{- 1} (cot x) = x, x \in (0, π) an d tan^{- 1} (- x) = - tan^{- 1} x ⎦ ⎤

= - \frac{π}{6} + \frac{π}{3} - \frac{π}{4} = \frac{- 2 π + 4 π - 3 π}{12}

= \frac{- 5 π + 4 π}{12} = \frac{- π}{12}

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Question Text	Find the value of $tan^{- 1} (- \frac{1}{3}) + cot^{- 1} (- \frac{1}{3}) + tan^{- 1} [sin (- \frac{- π}{2})]$ .
Answer Type	Text solution:1
Upvotes	150

Find the value of tan−1(−3​1​)+cot−1(−3​1​)+tan−1[sin(−2−π​)].

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Find the value of $tan^{- 1} (- \frac{1}{3}) + cot^{- 1} (- \frac{1}{3}) + tan^{- 1} [sin (- \frac{- π}{2})]$ .