If sin ⁡ ( α + β ) = 6 sin ⁡ ( α − β ) and k = tan ⁡ α tan ⁡ β , then, what is the value of k ?

Explanation: Given, sin ⁡ ( α + β ) = 6 sin ⁡ ( α − β ) and k = tan ⁡ α tan ⁡ β sin ⁡ ( α + β ) = 6 sin ⁡ ( α − β ) ⇒ sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β = 6 sin ⁡ α cos ⁡ β − 6 cos ⁡ α sin ⁡ β ⇒ cos ⁡ α sin ⁡ β + 6 cos ⁡ α sin ⁡ β = 6 sin ⁡ α cos ⁡ β − sin ⁡ α cos ⁡ β ⇒ 7 cos ⁡ α sin ⁡ β = 5 sin ⁡ α cos ⁡ β ⇒ sin ⁡ α cos ⁡ α = 7 5 × sin ⁡ β cos ⁡ β ⇒ tan α = 7 5 × tan ⁡ β ⇒ k = tan ⁡ α tan ⁡ β = 7 5 ∴ The value of k is 7 5 Hence, the correct option is (D).

Question:

If $\sin (α + β) = 6 \sin (α - β)$ and $k = \frac{\tan α}{\tan β}$ , then, what is the value of $k$ ?

MHT CET

Updated On: Aug 3, 2024

(A) $\frac{3}{5}$
(B) $\frac{6}{5}$
(C) $\frac{4}{5}$
(D) $\frac{7}{5}$

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The Correct Option is D

Solution and Explanation

Explanation:
Given,

\sin (α + β) = 6 \sin (α - β) and k = \frac{\tan α}{\tan β}

\sin (α + β) = 6 \sin (α - β)

\Rightarrow \sin α \cos β + \cos α \sin β = 6 \sin α \cos β - 6 \cos α \sin β

\Rightarrow \cos α \sin β + 6 \cos α \sin β = 6 \sin α \cos β - \sin α \cos β

\Rightarrow 7 \cos α \sin β = 5 \sin α \cos β

\Rightarrow \frac{\sin α}{\cos α} = \frac{7}{5} \times \frac{\sin β}{\cos β}

\Rightarrow tan α = \frac{7}{5} \times \tan β

\Rightarrow k = \frac{\tan α}{\tan β} = \frac{7}{5}

∴

The value of

k is \frac{7}{5}

Hence, the correct option is (D).

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