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

The correct option is A tan( α +β #160; 2) = b a sinα #160; +sinβ #160; =a cosα #160; cosβ #160; =b ∴sinα #160; +sinβ ...

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Basic Inverse Trigonometric Functions

If sinα + s...

Question

If $sin α + sin β = a$ and $cos α - cos β = b$ then

tan (\frac{α + β}{2}) = - \frac{b}{a}

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tan (\frac{α + β}{2}) = \frac{b}{a}

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cos (α + β) = \frac{a^{2} + b^{2} - 2}{2}

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cos (α + β) = \frac{2 - a^{2} - b^{2}}{2}

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Similar questions

\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 $s i n α + s i n β = α$ 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

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

\cos (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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