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

We know that sin^ 1x +cos^ 1x=π2 ∀ x ∈ R ⋯(1) Also, π2≤sin^ 1x ≤π2 ⋯(2) From (1) & (2), 0 ≤ 2 sin^ 1x + cos^ 1x ≤π nbsp; ...

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Polar Representation of a Complex Number

If α≤ 2 sin -...

Question

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

α = \frac{- π}{2}, β = \frac{π}{2}

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α = \frac{- π}{2}, β = \frac{3 π}{2}

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α = 0, β = π

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α = 0, β = 2 π

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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

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

| 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

α = {cos}^{- 1} (\frac{3}{5}), β = {tan}^{- 1} (\frac{1}{3}),

0 < α, β < \frac{π}{2},

α - β

cot α = \frac{1}{2}, s e c β = - \frac{5}{3}

π < α < \frac{3 π}{2}

\frac{π}{2} < β < π

tan (α + β)

α + β

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