Sin -1 a -sin -1-sin -1 y -3 -2- then a - ay - y is equal toA- 1B- 0C- 3D- -3

The correct option is C 3Since, sin, π/2≤ sin^ 1x≤π/2, sin^ 1a=π/2,sin^ 1β=π/2,sin^ 1y=π/2 ∴ a=β =y=1. Thus, aβ+ay+β y=3 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Range of Trigonometric Expressions

sin -1 a +sin...

Question

$s i n^{- 1} a + s i n^{- 1} β + s i n^{- 1} y = \frac{3 π}{2}$ ,then $a β + a y + β y$ is equal to

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 3
Since,
$s i n, - \frac{π}{2} \leq s i n^{- 1} x \leq \frac{π}{2},$
$s i n^{- 1} a = \frac{π}{2}, s i n^{- 1} β = \frac{π}{2}, s i n^{- 1} y = \frac{π}{2}$
$∴ a = β = y = 1$ . Thus, $a β + a y + β y = 3$

Suggest Corrections

Similar questions

If $s i n^{- 1} x + s i n^{- 1} y = \frac{π}{2}$ ,then $c o s^{- 1} x + c o s^{- 1} y$ is equal to

Q. If

{sin}^{- 1} α + {sin}^{- 1} β + {sin}^{- 1} γ = \frac{3 π}{2}

, then

α β + α γ + β γ

is equal to :

Q. lf

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

, then

x^{100} + y^{100} + z^{100} - \frac{9}{x^{101} + y^{101} + z^{101}} =

$s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = \frac{3 π}{2},$ then the value of $x^{100} + y^{100} + z^{100} - \frac{9}{x^{101} + y^{101} + z^{101}} =$

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

Join BYJU'S Learning Program

sin−1a+sin−1β+sin−1y=3π2,then aβ+ay+βy is equal to

The correct option is B 3Since, sin,−π2≤sin−1x≤π2, sin−1a=π2,sin−1β=π2,sin−1y=π2 ∴a=β=y=1. Thus, aβ+ay+βy=3

$s i n^{- 1} a + s i n^{- 1} β + s i n^{- 1} y = \frac{3 π}{2}$ ,then $a β + a y + β y$ is equal to

The correct option is B 3
Since,
$s i n, - \frac{π}{2} \leq s i n^{- 1} x \leq \frac{π}{2},$
$s i n^{- 1} a = \frac{π}{2}, s i n^{- 1} β = \frac{π}{2}, s i n^{- 1} y = \frac{π}{2}$
$∴ a = β = y = 1$ . Thus, $a β + a y + β y = 3$