If x - 0 and - - sin - 1x - cos - 1x - tan - 1x - then

The correct option is C 0 ≤θ≤π/2 Given x≥ 0tan^ 1 x∈[0,π2) tan^ 1 x∈( π2,0] θ=sin^ 1 x+cos^ 1 x tan^ 1 x=π2 tan^ 1 x As tan^ 1 x∈( π2,0] ...

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Formation of a Differential Equation from a General Solution

If x ≥ 0 an...

Question

If $x \geq 0$ and $θ = {sin}^{- 1} x + {cos}^{- 1} x - {tan}^{- 1} x$ , then

\frac{π}{2} \leq θ \leq \frac{3 π}{4}

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0 \leq θ \leq \frac{π}{4}

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0 \leq θ \leq \frac{π}{2}

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\frac{π}{4} \leq θ \leq \frac{π}{2}

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If $θ = {sin}^{- 1} x + {cos}^{- 1} x - {tan}^{- 1} x, x \geq 0$ then the smallest interval in which $θ$ lies is

\frac{π}{4}

θ

\frac{π}{4}

θ

θ

tan [\frac{π}{4} + θ] + tan [\frac{π}{4} - θ] = α

{tan}^{3} [\frac{π}{4} + θ] + {tan}^{3} [\frac{π}{4} - θ] =

θ, 0 < θ < \frac{π}{2},

6 \sum m = 1 c o s e c [θ + \frac{(m - 1) π}{4}] c o s e c [θ + \frac{m π}{4}] = 4 \sqrt{2}

\frac{π}{12}

\frac{5 π}{12}

tan θ + cot θ = 4, 0 < θ < \frac{π}{2},

θ = \frac{π}{12}

\frac{5 π}{12}

Which of the following statements are correct?

1 . if sin θ = sin α ⇒ θ = nπ + $(- 1)^{n}$ α where α ∈ [ - $\frac{π}{2}$ , $\frac{π}{2}$ ] n ∈ I

2 . if cos θ = cos α ⇒ 2nπ±α where α ∈ [0,π] n ∈ I

3 . if tan θ = tan α ⇒ θ = nπ + α where α ∈ (- $\frac{π}{2}$ , $\frac{π}{2}$ ) n ∈ I

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