If sin-sin- a andcos-cos- b- then the value of sin - is equal toA- abB- a-bC- 2 a b-a2-b2D- 2 a b-a2-b2

The correct option is C 2aba^2+b^2 nbsp; nbsp;sinα + sinβ =a ⇒ 2 sinα+ β2.cosα β2=a ⋯(1) cosα +cosβ=b ⇒ 2 cosα+ β2.cosα β2=b ⋯(2) Dividing (1) an ...

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Trigonometric Ratios of Allied Angles

If sinα+sinβ=...

Question

$If sin α + sin β = a$ and
$cos α + cos β = b$ , then the value of $sin (α + β)$ is equal to

a b

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a + b

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

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

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s i n α + s i n β = a

c o s α + c o s β = b

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

sin α + sin β = a

cos α + cos β = b

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

m

\sin α + \sin β = a and \cos α - \cos β = b then \tan \frac{α - β}{2} =

- \frac{a}{b}

- \frac{b}{a}

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

\frac{cos α}{cos β} = a

\frac{sin α}{sin β} = b

(a^{2} - b^{2}) {sin}^{2} β = a^{2} - 1

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