If alpha - 2 sin -1 x -cos -1 x - thenA- -2-3 -2B- -0 - -0 -2 -D- -2-2

The correct option is B α =0 β = π α≤ sin^ 1x+π/2≤β sin^ 1x ϵ [ π/2,π/2 nbsp; ]⇒α =0,β =π ...

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

Sign of Trigonometric Ratios in Different Quadrants

If alpha ≤ 2 ...

Question

If $a l p h a \leq 2 s i n^{- 1} x + c o s^{- 1} x \leq β$ , then

$α = \frac{- π}{2} β = \frac{π}{2}$

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

$α = \frac{- π}{2} β = \frac{3 π}{2}$

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

$α = 0 β = π$

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

$α = 0 β = 2 π$

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

Open in App

Solution

Suggest Corrections

Similar questions

α \leq 2 {sin}^{- 1} x + {cos}^{- 1} x \leq β,

α, β (α > β)

{sin}^{- 1} x - \frac{1}{{sin}^{- 1} x} = {cos}^{- 1} x - \frac{1}{{cos}^{- 1} x} .

If α \leq 2 \sin^{- 1} x + \cos^{- 1} x \leq β, then (a) α = - \frac{π}{2}, β = \frac{π}{2} (b) α = 0, β = π (c) α = - \frac{π}{2}, β = \frac{3 π}{2} (d) α = 0, β = 2 π

| s i n^{- 1} x | + | t a n^{- 1} x | + | c o s^{- 1} x | = | π - c o t^{- 1} x |

[α, β]

α a n d β

tan α = \frac{1}{7}

tan β = \frac{1}{\sqrt{10}}

α + 2 β = \frac{π}{a}

0 < α < \frac{π}{2}

0 < β < \frac{π}{2}

a

Join BYJU'S Learning Program