If sin- - sin- - a and cos- - cos- - b- then tan- - -2 -

The correct option is B ba Solution : sinα +sinβ #160; = a and cosα cosβ #160; = b ⇒ 2 sin α +β2.sinα β2 = a ...(1) and 2sinα ...

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Sin2A and Cos2A in Terms of tanA

If sinα + s...

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

- \frac{a}{b}

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- \frac{b}{a}

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

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none of these

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If $s i n α + s i n β = α$ and $c o s α - c o s β = b$ , then $t a n \frac{α - β}{2} =$

\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}}

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

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

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}}

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