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Question
`sin^-1(1 - x) - 2sin^-1 x = pi/2`, tan 'x' is equal to
Options
`0, 1/2`
`1, 1/2`
0
`1/2`
MCQ
Solution
0
Explanation:
`sin^-1(1 - x) - 2sin^-1 x = pi/2`
⇒ `- 2 sin x = pi/2 - sin^-1 = 2pi(4 - x)`
⇒ `- 2sin^-1x = cos^-1(1 - x)`
Let `sin^-1x` = θ
⇒ sin θ = x
⇒ θ = `cos^-1 (sqrt(1 - x^2))`
∴ `sin^-1x = cos^-1 (sqrt(1 - x^2))`
From equation (1) we have
`- 2cos^-1 (sqrt(1 - x^2)) = cos^-1(1 - x)`
Put x = sin y
`- 2cos^-1 (sqrt(1 - sin^2y)) = cos^1(1 - sin y)`
⇒ `- 2cos^-1 (cos y) = cos^-1 (1 - sin y)`
⇒ `- 2y = cos^-1(1 - sin y)`
⇒ `1 - sin y = cos (- 2y) = cos 2y`
⇒ `1 - sin y = 1 - sin^2y`
⇒ `2sin^2y - sin y` = 0
⇒ `sin y (2 sin y - 1)` = 0
⇒ `sin y` = 0 or `1/2`
∴ x = 0 or x = `1/2`
But when x = `1/2`, does not satisfy the equation, Thus x = 0 is the only solution.
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