Evaluate: tan−1[2cos(2sin−1(21))]=aπ. Find a

Evaluate : tan−1[2cos(2sin−1(21))]so, tan−1[2cos(2sin−121)]=tan−1[2cos(2(6π))]=tan−1[2cos3π]=tan−1[2.21]=tan−1(1)=4πso, tan−1[2cos(2sin−1(21))]=4πso, a=4

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Evaluate: ^{-1}[2 (2^{-1}({1}{2}))]={}{a}. Find a

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Evaluate: $tan^{- 1} [2 cos (2 sin^{- 1} (\frac{1}{2}))]$ = $\frac{π}{a}$ . Find a

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Evaluate:

tan^{- 1} [2 cos (2 sin^{- 1} (\frac{1}{2}))]

\frac{π}{a}

. Find a

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Evaluate :

t a n^{- 1} [2 cos (2 s i n^{- 1} (\frac{1}{2}))]

so,

t a n^{- 1} [2 cos (2 s i n^{- 1} \frac{1}{2})]

= t a n^{- 1} [2 cos (2 (\frac{π}{6}))]

= t a n^{- 1} [2 cos \frac{π}{3}]

= t a n^{- 1} [2. \frac{1}{2}]

= t a n^{- 1} (1)

= \frac{π}{4}

so,

t a n^{- 1} [2 cos (2 s i n^{- 1} (\frac{1}{2}))] = \frac{π}{4}

so,

a = 4

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Question Text	Evaluate: $tan^{- 1} [2 cos (2 sin^{- 1} (\frac{1}{2}))]$ = $\frac{π}{a}$ . Find a
Answer Type	Text solution:1
Upvotes	150

Evaluate: tan−1[2cos(2sin−1(21​))]=aπ​. Find a

Text solutionVerified

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Evaluate: $tan^{- 1} [2 cos (2 sin^{- 1} (\frac{1}{2}))]$ = $\frac{π}{a}$ . Find a