If sin-sin- and cos-cos-b- then tan-2-A- None of theseB- -5C-D- -a2-b2

The correct option is Given: sin α +sinβ =a …(i) cos α cosβ =b …(ii) Dividing (i) by(ii) ⇒2sin ( α +β/2 )cos ( α β/2 )/ 2sin ( α +β/2 )cos ( α β/ ...

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Trigonometric Equations

If sinα+sinβ=...

Question

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

$- \frac{a}{b}$

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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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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}}

sin α + sin β = a

cos α - cos β = b,

tan \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}}

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