If sin -cos - then which of the following is-are correct- where - -0- -2-4

The correct options are A tanα = 1 tanθ1+tanθ D tanθ nbsp; = 1 tanα1 +tanαsin (θ +α) = cos (θ +α) ⇒sinθcosα +sinαcosθ nbsp; = cosθcosα sinθsinα ⇒s ...

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Byju's Answer

Standard XII

Mathematics

Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

If sin θ+α=co...

Question

If $sin (θ + α) = cos (θ + α)$ , then which of the following is/are correct?
where $θ, α \in (0, \frac{π}{2}) - {\frac{π}{4}}$

tan α = \frac{1 - tan θ}{1 + tan θ}

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tan α = \frac{1 + tan θ}{1 - tan θ}

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tan θ = \frac{1 + tan α}{1 - tan α}

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tan θ = \frac{1 - tan α}{1 + tan α}

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

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Solution

The correct options are
A $tan α = \frac{1 - tan θ}{1 + tan θ}$
D $tan θ = \frac{1 - tan α}{1 + tan α}$
$sin (θ + α) = cos (θ + α) \Rightarrow sin θ cos α + sin α cos θ = cos θ cos α - sin θ sin α \Rightarrow sin θ cos α + sin θ sin α = cos θ cos α - sin α cos θ \dots (1) \Rightarrow sin θ (cos α + sin α) = cos θ (cos α - sin α) \Rightarrow tan θ = \frac{(cos α - sin α)}{(cos α + sin α)} \Rightarrow tan θ = \frac{1 - tan α}{1 + tan α}$

Now,
$tan θ = \frac{1 - tan α}{1 + tan α} \Rightarrow tan θ + tan α tan θ = 1 - tan α \Rightarrow tan α = \frac{1 - tan θ}{1 + tan θ}$

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Similar questions

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

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

Q. If

cos (θ - α) = a

and

sin (θ - β) = b

(0 < θ - α, θ - β < \frac{π}{2})

, then prove that

{cos}^{2} (α - β) + 2 a b sin (α - β)

is equal to

a^{2} - b^{2}

Q. If the tangent to the curve

x = a (θ + sin θ), y = a (1 + cos θ)

θ = \frac{π}{3}

makes an angle

α (0 \leq α < π)

with x-axis, then

α

Q. Consider three points

P = [- sin (β - α), - cos β], Q = [cos (β - α), sin β]

and

R = [cos (β - α + θ), sin (β - θ)],

where

0 < α, β, θ < \frac{π}{4} .

Then

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If sin(θ+α)=cos(θ+α), then which of the following is/are correct? where θ,α∈(0,π2)−{π4}

If $sin (θ + α) = cos (θ + α)$ , then which of the following is/are correct?
where $θ, α \in (0, \frac{π}{2}) - {\frac{π}{4}}$