If tan- -cos- -sin-cos- -sin- and - is acute then - -

The correct option is D π 4 tanθ=cosα sinαcosα+sinα =cosα(1 tanα)cosα(1+tanα) =1 tanα(1+tanα =tanπ4 tanα(1+tanπ4tanα =tan(π4 α) ...

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Integration by Substitution

If tanθ =co...

Question

If $tan θ = \frac{cos α - sin α}{cos α + sin α}$ and $θ$ is acute then $θ + α$ =

\frac{π}{5}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

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lim α \to π / 4 \frac{sin α - cos α}{α - π / 4}

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

\frac{sin α}{sin β} = \frac{\sqrt{3}}{2}

\frac{cos α}{cos β} = \frac{\sqrt{5}}{2}, 0 < α, β < \frac{π}{2}

0 < α, β < \frac{π}{4}, cos (α + β) = \frac{4}{5}, sin (α - β) = \frac{5}{13}

tan 2 α =

cos (α + β) = \frac{3}{5}, sin (α - β) = \frac{5}{13}

0 < α, β < \frac{π}{4},

tan (2 α)

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