Cos - - sin - - -cos - -sin- - -cos - -sin- - is equal to

The correct option is A 0Given Expression: cos #160;α sin(β γ )+cos #160;β #160; #160;sin(γ α )+cos #160;γ #160;sin(α β ) ⇒si ...

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Standard XII

Mathematics

Principal Solution of Trigonometric Equation

cos α . sin ...

Question

$c o s α . s i n (β - γ) + c o s β . s i n (γ - α) + c o s γ . s i n (α - β)$ is equal to

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4 c o s α c o s β c o s γ

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Solution

The correct option is A 0
Given Expression: $c o s α s i n (β - γ) + c o s β s i n (γ - α) + c o s γ s i n (α - β)$

$\Rightarrow sin (A - B) = sin A cos B - sin B cos A$

On Expanding the cosine terms we get

$= c o s α s i n β c o s γ - c o s α c o s γ s i n γ + c o s β s i n γ c o s α - c o s β c o s γ s i n α + c o s γ s i n α s i n β$

$= 0$

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cosα.sin(β−γ)+cosβ.sin(γ−α)+cosγ.sin(α−β) is equal to

The correct option is A 0Given Expression: cosα sin(β−γ)+cosβsin(γ−α)+cosγsin(α−β)⇒sin(A−B)=sinAcosB−sinBcosAOn Expanding the cosine terms we get=cosαsinβcosγ−cosαcosγsinγ+cosβsinγcosα−cosβcosγsinα+cosγsinαsinβ=0

$c o s α . s i n (β - γ) + c o s β . s i n (γ - α) + c o s γ . s i n (α - β)$ is equal to