If α=(1)/(2) (x+(1)/(x)) and β=(1)/(2) left(y+(1)/(y)),( x y 0) x, y, α, β ∈ R then

Correct answer: D. (α+β+γ)= α+ β+ γ α, β, γ ∈ R aligned & α=(1)/(2)(x+(1)/(x)) and β=(1)/(2)(y+(1)/(y)) & since x y 0 x+(1)/(x) ≥ 2 or ≤-2 & y+(1)/(y) ≥ 2 or ≤-2 aligned α=1, β=1 ....(1) or α=-1, β=-1 ...(2) aligned & α β=1 & ( α+ β)^2=4 & α+β is an even multiple of π & (α+β+γ)= (2 n π+γ)= γ & Also, α= β=0 aligned

JEE Advanced

Exam: JEE Advanced
Subject: Mathematics
Chapter: Trigonometry

JEE Advanced · Mathematics · Trigonometry

If α=(1)/(2) (x+(1)/(x)) and β=(1)/(2) left(y+(1)/(y)),( x y 0) x,...

\cos \alpha=\frac{1}{2}

\left(x+\frac{1}{x}\right)

and \cos \beta=\frac{1}{2}\l

left(y+\frac{1}{y}),(

x y\gt0) x, y, \alpha, \beta \in R

then

Options

A $\sin (\alpha+\beta+\gamma)=\sin \gamma \forall \gamma \in \mathrm{R}$
B $\cos \alpha \cos \beta=1 \forall \alpha, \beta \in \mathrm{R}$
C $(\cos \alpha+\cos \beta)^2=4 \forall \alpha, \beta \in R$
D $\sin (\alpha+\beta+\gamma)=\sin \alpha+\sin \beta+\sin \gamma \forall \alpha, \beta, \gamma \in \mathrm{R}$

Correct answer

D. $\sin (\alpha+\beta+\gamma)=\sin \alpha+\sin \beta+\sin \gamma \forall \alpha, \beta, \gamma \in \mathrm{R}$

Step-by-step solution

\begin{aligned} & \cos \alpha=\frac{1}{2}\left(x+\frac{1}{x}\right) \text { and } \cos \beta=\frac{1}{2}\left(y+\frac{1}{y}\right) \\ & \text { since } x y\gt0 \Rightarrow x+\frac{1}{x} \geq 2 \text { or } \leq-2 \\ & y+\frac{1}{y} \geq 2 \text { or } \leq-2\end{aligned}

\Rightarrow \cos \alpha=1, \cos \beta=1

....(1) or

\cos \alpha=-1, \cos \beta=-1

...(2)

\begin{aligned} & \therefore \cos \alpha \cos \beta=1 \\ & (\cos \alpha+\cos \beta)^2=4 \\ & \Rightarrow \alpha+\beta \text { is an even multiple of } \pi \\ & \Rightarrow \sin (\alpha+\beta+\gamma)=\sin (2 n \pi+\gamma)=\sin \gamma \\ & \text { Also, } \sin \alpha=\sin \beta=0\end{aligned}

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