If sin -11-3-sin -12-3-sin -1 x- then x is equal to-Roorkee 1995-A- 0B- -5-4 -2-9C- -5-4 -2-9D-

The correct option is C √(5)+4√(2)/9 sin^ 11/3 + sin^ 12/3 = sin^ 1 [ 1/3√(1 4/9) + 2/3√(1 1/9) ] = sin^ 1 [ √(5) + 4√(2)/9 ] Therefore x = √(5)+4√(2)/9 ...

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Standard XI

Mathematics

Trigonometric Ratios for Sum of Two Angles

If sin -11/3+...

Question

If $s i n^{- 1} \frac{1}{3} + s i n^{- 1} \frac{2}{3} = s i n^{- 1} x,$ then x is equal to
[Roorkee 1995]

0.0

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\frac{\sqrt{5} - 4 \sqrt{2}}{9}

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$\frac{\sqrt{5} + 4 \sqrt{2}}{9}$

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$\frac{x}{2}$

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Solution

The correct option is C
$\frac{\sqrt{5} + 4 \sqrt{2}}{9}$

$s i n^{- 1} \frac{1}{3} + s i n^{- 1} \frac{2}{3} = s i n^{- 1} [\frac{1}{3} \sqrt{1 - \frac{4}{9}} + \frac{2}{3} \sqrt{1 - \frac{1}{9}}] = s i n^{- 1} [\frac{\sqrt{5} + 4 \sqrt{2}}{9}]$
Therefore $x = \frac{\sqrt{5} + 4 \sqrt{2}}{9}$

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If sin−113+sin−123=sin−1 x, then x is equal to [Roorkee 1995]

The correct option is C √5+4√29 sin−113+sin−123=sin−1[13√1−49+23√1−19]=sin−1[√5+4√29] Therefore x=√5+4√29

If $s i n^{- 1} \frac{1}{3} + s i n^{- 1} \frac{2}{3} = s i n^{- 1} x,$ then x is equal to
[Roorkee 1995]

The correct option is C
$\frac{\sqrt{5} + 4 \sqrt{2}}{9}$

$s i n^{- 1} \frac{1}{3} + s i n^{- 1} \frac{2}{3} = s i n^{- 1} [\frac{1}{3} \sqrt{1 - \frac{4}{9}} + \frac{2}{3} \sqrt{1 - \frac{1}{9}}] = s i n^{- 1} [\frac{\sqrt{5} + 4 \sqrt{2}}{9}]$
Therefore $x = \frac{\sqrt{5} + 4 \sqrt{2}}{9}$