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

Sinα/a=cosα/b⇒ tan α =a/b a sin 2α +b cos 2α =a2tanα/1+tan^2α+b1 tan^2α/1+tan^2α=a.2ab/a^2+b^2+b. ( b^2 a^2/b^2+a^2 ) =b ( 2a^2+b^2 a^2 )/a^2+b^2=b. ...

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Standard XIII

Mathematics

Sin2A and Cos2A in Terms of tanA

If sinα/a=cos...

Question

If $\frac{s i n α}{a} = \frac{c o s α}{b}, t h e n a s i n 2 α + b c o s 2 α$ =

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a+b

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Solution

The correct option is B b
$\frac{s i n α}{a} = \frac{c o s α}{b} \Rightarrow t a n α = \frac{a}{b}$
$a sin 2 α + b c o s 2 α = a \frac{2 t a n α}{1 + t a n^{2} α} + b \frac{1 - t a n^{2} α}{1 + t a n^{2} α} = a . \frac{2 a b}{a^{2} + b^{2}} + b . (\frac{b^{2} - a^{2}}{b^{2} + a^{2}})$
$= \frac{b (2 a^{2} + b^{2} - a^{2})}{a^{2} + b^{2}} = b$ .

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If sinαa=cosαb, then a sin 2α+b cos 2α =

The correct option is B bsinαa=cosαb⇒tan α=ab a sin 2α+b cos 2α=a2tanα1+tan2α+b1−tan2α1+tan2α=a.2aba2+b2+b.(b2−a2b2+a2) =b(2a2+b2−a2)a2+b2=b.

If $\frac{s i n α}{a} = \frac{c o s α}{b}, t h e n a s i n 2 α + b c o s 2 α$ =