If cos x-a-sin x-b then -a cos 2 x-b sin 2 x- is equal to

The correct option is B |a| cosx/a = sinx/b⇒ tanx = b/a ⇒ cos^2x = 1 b^2/a^2/1 + nbsp;b^2/a^2 = a^2 b^2/a^2 + nbsp;b^2 ⇒ acos^2x = aa(a^2 b^2)/a^2 + ...

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Byju's Answer

Standard XII

Mathematics

Inequalities of Integrals

If cos x/a=si...

Question

If $\frac{c o s x}{a} = \frac{s i n x}{b}$ then |acos2x + bsin2x| is equal to

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|a|

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Solution

The correct option is D
|a|

$\frac{c o s x}{a} = \frac{s i n x}{b} \Rightarrow t a n x = \frac{b}{a}$

$\Rightarrow c o s^{2} x = \frac{1 - \frac{b^{2}}{a^{2}}}{1 + \frac{b^{2}}{a^{2}}} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$

$\Rightarrow a c o s^{2} x = a \frac{a (a^{2} - b^{2})}{a^{2} + b^{2}} \Rightarrow s i n^{2} x = \frac{2 \frac{b}{a}}{1 + \frac{b^{2}}{a^{2}}} = \frac{2 a b}{a^{2} + b^{2}}$

$\Rightarrow b s i n 2 x = \frac{2 a b^{2}}{a^{2} + b^{2}}$

$\Rightarrow$ |acos 2x + bsin 2x| = $\frac{a^{3} + a b^{2}}{a^{2} + b^{2}} = | a |$

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If $\frac{c o s x}{a} = \frac{s i n x}{b}$ then $| a c o s 2 x + b s i n 2 x | =$

Q. If

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tan x = \frac{b}{a}

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If cosxa=sinxb then |acos2x + bsin2x| is equal to

The correct option is D |a| cosxa=sinxb⇒tanx=ba ⇒cos2x=1−b2a21+b2a2=a2−b2a2+b2 ⇒acos2x=aa(a2−b2)a2+b2⇒sin2x=2ba1+b2a2=2aba2+b2 ⇒bsin2x=2ab2a2+b2 ⇒ |acos 2x + bsin 2x| = a3+ab2a2+b2=|a|

If $\frac{c o s x}{a} = \frac{s i n x}{b}$ then |acos2x + bsin2x| is equal to