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Sign of Trigonometric Ratios in Different Quadrants

The sinα+si...

Question

The $sin α + sin β = a$ and $cos α + cos β = b$ , then the value of $sin (α - β)$ is

A

\frac{2 a b}{(a^{2} - b^{2})}

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B

\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}

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C

\frac{2 a b}{a^{2} + b^{2}}

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D

N o n e o f t h e s e

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Solution

The correct option is C $\frac{2 a b}{a^{2} + b^{2}}$
$a=sin\alpha +sin\beta$
$b=cos\alpha +cos\beta$
$a^{2}+b^{2}=4cos^{2}\frac{\alpha -\beta}{2}$
$(a+b)^{2}=4cos^{2}\frac{\alpha-\beta}{2} + 8.cos^{2}\frac{\alpha-\beta}{2}.sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$a^{2}+b^{2}+2ab=4cos^{2}\frac{\alpha-\beta}{2} + 8.cos^{2}\frac{\alpha-\beta}{2}.sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$4cos^{2}\frac{\alpha-\beta}{2}+2ab=4cos^{2}\frac{\alpha-\beta}{2} + 8.cos^{2}\frac{\alpha-\beta}{2}.sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$2ab=24.cos^{2}\frac{\alpha-\beta}{2}.sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$2ab=(a^{2}+b^{2})2sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$\frac{2ab}{a^{2}+b^{2}}=2sin\frac{\alpha+\beta}{2}.cos\frac{\alpha-\beta}{2}$
$\frac{2ab}{a^{2}+b^{2}}=sin(\alpha +\beta )$
hence the value of $sin(\alpha +\beta )=\frac{2ab}{a^{2}+b^{2}}$

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Similar questions

s i n α + s i n β = a

c o s α + c o s β = b,

s i n (α + β) = \frac{2 a b}{a^{2} + b^{2}}

c o s (α + β) = \frac{b^{2} - a^{2}}{b^{2} + a^{2}}

If $s i n α + s i n β = a a n d c o s α + c o s β = b$ , prove that

(i) $s i n (α + β) = \frac{2 a b}{a^{2} + b^{2}}$

(ii) $c o s (α - β) = \frac{a^{2} + b^{2} - 2}{2}$

If sin α + sin β = a

cos α + cos β = b

, then the value of

sin (α + β)

sin α + sin β = a

cos α + cos β = b

, then the value of

sin α + β = \frac{m a b}{a^{2} + b^{2}}

m

sin α + sin β = a

cos α + cos β = b

tan \frac{α - β}{2} = \pm \sqrt{\frac{x - a^{2} - b^{2}}{a^{2} + b^{2}}}

x

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