If α, β, γ ∈[0, π] and if α, β, γ are in AP, then ( α- γ)/( γ- α) is equal to

Correct answer: C. β ( α- γ)/( γ- α)=(2 (α+γ)/(2) (α-γ)/(2))/(2 (α+γ)/(2) · (α-γ)/(2))= (α+γ)/(2)[ array l A- B=2 ((A+B)/(2)) ((A-B)/(2)) and A- B=2 ((A+B)/(2)) ((B-A)/(2)) array ] But α, β, γ are in AP (α+γ)/(2)=β So, required value is β.

BITSAT 2022

Exam: BITSAT
Year: 2022
Subject: Mathematics
Chapter: Sequences and Series

BITSAT 2022 · Mathematics · Sequences and Series

If α, β, γ ∈[0, π] and if α, β, γ are in AP, then ( α- γ)/( γ- α)...

\alpha, \beta, \gamma \in[0, \pi]

and if

\alpha, \beta, \gamma

are in AP, then

\frac{\sin \alpha-\sin \gamma}{\cos \gamma-\cos \alpha}

is equal to

Options

A $\sin \beta$
B $\cos \beta$
C $\cot \beta$
D $2 \cos \beta$

Correct answer

C. $\cot \beta$

Step-by-step solution

\frac{\sin \alpha-\sin \gamma}{\cos \gamma-\cos \alpha}=\frac{2 \cos \frac{\alpha+\gamma}{2} \sin \frac{\alpha-\gamma}{2}}{2 \sin \frac{\alpha+\gamma}{2} \cdot \sin \frac{\alpha-\gamma}{2}}=\cot \frac{\alpha+\gamma}{2}\left[

\begin{array}{l}\because \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right) \\ \text{ and } \cos A-\cos B=2 \sin \left(\frac{A+B}{2}\right) \sin\left(\frac{B-A}{2}\right)\end{array}

\right] But \alpha, \beta, \gamma are in

AP\Rightarrow \quad

\frac{\alpha+\gamma}{2}=\beta So, required value is

\cot

\beta.

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