If sin-1x -sin-1y -sin-1z -3-2then -r-12x100r-y103r- x201y201 -

We know that π/2≤sin^ 1x≤π/2 π/2≤sin^ 1y≤π/2 π/2≤sin^ 1z≤π/2 ∴ sin^ 1x+sin^ 1y+sin^ 1z=3π/2 ⇒ sin^ 1x=sin^ 1y=sin^ 1z=π/2 ⇒ x=y=z=1 ∴∑^2_r=1(x^100r+y^103r) ...

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Byju's Answer

Standard XI

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

If sin-1x +si...

Question

If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2}$
then $\frac{\sum_{r = 1}^{2} (x^{100 r} + y^{103 r})}{\sum x^{201} y^{201}}$ =

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\frac{4}{3}

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Solution

The correct option is D $\frac{4}{3}$
We know that $- \frac{π}{2} \leq {sin}^{- 1} x \leq \frac{π}{2} - \frac{π}{2} \leq {sin}^{- 1} y \leq \frac{π}{2} - \frac{π}{2} \leq {sin}^{- 1} z \leq \frac{π}{2} ∴ s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2} \Rightarrow s i n^{- 1} x = s i n^{- 1} y = s i n^{- 1} z = \frac{π}{2} \Rightarrow x = y = z = 1 ∴ \frac{\sum_{r = 1}^{2} (x^{100 r} + y^{103 r})}{\sum x^{201} y^{201}} = \frac{x^{100} + y^{103} + x^{200} + y^{206}}{x^{201} y^{201} + y^{201} z^{201} + z^{201} x^{201}} = \frac{1 + 1 + 1 + 1}{1.1 + 1.1 + 1.1} = \frac{4}{3}$

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Similar questions

Q. If

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

then

\frac{\sum_{r = 1}^{2} (x^{100 r} + y^{103 r})}{\sum x^{201} y^{201}}

Q. Assertion :If

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

,then

\frac{3 \sum_{r = 1}^{2008} (x^{r} + y^{r})}{2 \sum_{r = 1}^{2008} (x^{r} y^{r})} = 3

Reason:

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

is possible only if

x = y = z = 1

Q. If

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

, then xy +yx -xz =

Q. If

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

, then

(x^{500} + y^{500} + z^{500}) - (x^{501} + y^{501} + z^{501})

Q. If

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

, then find the minimum value of

x^{2} + y^{2} + z^{2}

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If sin−1x+sin−1y+sin−1z=3π2 then ∑2r=1(x100r+y103r)∑x201y201 =

The correct option is D 43We know that −π2≤sin−1x≤π2−π2≤sin−1y≤π2−π2≤sin−1z≤π2∴sin−1x+sin−1y+sin−1z=3π2⇒sin−1x=sin−1y=sin−1z=π2⇒x=y=z=1∴∑2r=1(x100r+y103r)∑x201y201=x100+y103+x200+y206x201y201+y201z201+z201x201=1+1+1+11.1+1.1+1.1=43

If $s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2}$
then $\frac{\sum_{r = 1}^{2} (x^{100 r} + y^{103 r})}{\sum x^{201} y^{201}}$ =