If sin−131+sin−132=sin−1x, then x is equal to-

Given sin−131+sin−132=sin−1xTo find x Sol :- sin−131+sin−132=sin−1(311−94+321−91) [sin−1x+sin−1y=sin−1(x+1−y2+y1−x2)]=sin−1(3135+3238)=sin−1(95+42)∵sin−131+sin−132=sin−1x (Given)∴x=95+42

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If ^{-1}{1}{3} + ^{-1}{2}{3} = ^{-1}x, then x is equal to-

Question

If $sin^{- 1} \frac{1}{3} + sin^{- 1} \frac{2}{3} = sin^{- 1} x$ , then $x$ is equal to-

0

\frac{5 - 4 2}{9}

\frac{5 + 4 2}{9}

\frac{π}{2}

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Emily Discussed

sin^{- 1} \frac{1}{3} + sin^{- 1} \frac{2}{3} = sin^{- 1} x

, then

x

is equal to-

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Given

s i n^{- 1} \frac{1}{3} + s i n^{- 1} \frac{2}{3} = s i n^{- 1} x

To find x

Sol :-

s i n^{- 1} \frac{1}{3} + s i n^{- 1} \frac{2}{3}

= s i n^{- 1} (\frac{1}{3} 1 - \frac{4}{9} + \frac{2}{3} 1 - \frac{1}{9})

[s i n^{- 1} x + s i n^{- 1} y = s i n^{- 1} (x + 1 - y^{2} + y 1 - x^{2})]

= s i n^{- 1} (\frac{1}{3} \frac{5}{3} + \frac{2}{3} \frac{8}{3})

= s i n^{- 1} (\frac{5 + 4 2}{9})

∵ s i n^{- 1} \frac{1}{3} + s i n^{- 1} \frac{2}{3} = s i n^{- 1} x

(Given)

∴ x = \frac{5 + 4 2}{9}

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Question Text	If $sin^{- 1} \frac{1}{3} + sin^{- 1} \frac{2}{3} = sin^{- 1} x$ , then $x$ is equal to-
Answer Type	Text solution:1
Upvotes	150

If sin−131​+sin−132​=sin−1x, then x is equal to-

Text solutionVerified

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If $sin^{- 1} \frac{1}{3} + sin^{- 1} \frac{2}{3} = sin^{- 1} x$ , then $x$ is equal to-