Sin -1-x -1-x-x-1-x2-

The correct option is C sin^ 1x sin^ 1√(x) sin^ 1 [ x√(1 x) √(x)√(1 x^2) ] Substitute x=sinθ and √(x)=sinϕ so that θ =sin^ 1x and ϕ =sin^ 1√(x) We get sin^ ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios for Sum of Two Angles

sin -1[x √1-x...

Question

$s i n^{- 1} [x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^{2}}] =$

$s i n^{- 1} x - s i n^{- 1} \sqrt{1 - x}$

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$s i n^{- 1} x + s i n^{- 1} \sqrt{1 - x}$

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$s i n^{- 1} x - s i n^{- 1} \sqrt{x}$

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Solution

The correct option is C
$s i n^{- 1} x - s i n^{- 1} \sqrt{x}$

$s i n^{- 1} [x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^{2}}]$
Substitute $x = s i n θ a n d \sqrt{x} = s i n ϕ s o t h a t θ = s i n^{- 1} x a n d ϕ = s i n^{- 1} \sqrt{x}$
We get $s i n^{- 1} [s i n θ \sqrt{1 - s i n^{2} ϕ} - s i n ϕ \sqrt{1 - s i n^{2} θ}]$
$= s i n^{- 1} [s i n θ c o s ϕ - s i n ϕ c o s θ] = s i n^{- 1} [s i n (θ - ϕ)] = θ - ϕ = s i n^{- 1} x - s i n^{- 1} \sqrt{x}$

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sin−1[x√1−x−√x√1−x2]=

The correct option is C sin−1x−sin−1√x sin−1[x√1−x−√x√1−x2] Substitute x=sinθ and √x=sinϕ so that θ=sin−1x and ϕ=sin−1√x We get sin−1[sinθ√1−sin2ϕ−sinϕ√1−sin2θ] =sin−1[sinθcosϕ−sinϕcosθ]=sin−1[sin(θ−ϕ)]=θ−ϕ=sin−1x−sin−1√x

$s i n^{- 1} [x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^{2}}] =$