Question

No. 20 please.

With explanation;

Answer 1

We have :

$\cos(2 \theta) = \frac{1 - \tan {}^{2} (\theta) }{1 + \tan {}^{2} (\theta)}$

$\sin(2 \theta) = \frac{2 \tan( \theta) }{1 + \tan {}^{2} (\theta)}$

✓_✓ Putting our values :

$b \cos(2 \theta) + \sin(2 \theta) = \frac{b -b {( \frac{a}{b})}^{2} + 2 (\frac{a}{b}) }{1 + { (\frac{a}{b}) }^{2} }$

-> Which simplifies to :

$\frac{ {b}^{3} - {a}^{2}b + 2ab}{ {a}^{2} + {b}^{2} }$

=> Answer = D

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Correct the Question to "b cos 2x + a sin2x " and then solve! You'll get the right answer then ^^"