If tan x-ba- then the value of a cos 2x-bsin 2x is-

The correct option is A a sin2x=2tan x1+tan^2x=2 ba1+( ba)^2=2aba^2+b^2 cos 2x=1 tan^2x1+tan^2x=1 ( ba)^21+( ba)^2=a^2 b^2a^2+b^2 Now, acos2x+bsin ...

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Standard XII

Mathematics

Parametric Differentiation

If tan x=ba...

Question

If $tan x = \frac{b}{a}$ , then the value of $a cos 2 x + b sin 2 x$ is?

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a - b

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Solution

The correct option is A a
$sin 2 x = \frac{2 tan x}{1 + {tan}^{2} x} = \frac{2 \frac{b}{a}}{1 + (\frac{b}{a})^{2}} = \frac{2 a b}{a^{2} + b^{2}} cos 2 x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2} x} = \frac{1 - (\frac{b}{a})^{2}}{1 + (\frac{b}{a})^{2}} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$

Now, $a cos 2 x + b sin 2 x = a . \frac{(a^{2} - b^{2})}{a^{2} + b^{2}} + b \frac{2 a b}{a^{2} + b^{2}} = \frac{a^{3} - a b^{2} + 2 a b^{2}}{a^{2} + b^{2}} = \frac{a (a^{2} + b^{2})}{a^{2} + b^{2}} = a$

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If tanx=ba, then the value of acos2x+bsin2x is?

The correct option is A asin2x=2tanx1+tan2x=2ba1+(ba)2=2aba2+b2cos2x=1−tan2x1+tan2x=1−(ba)21+(ba)2=a2−b2a2+b2Now, acos2x+bsin2x=a.(a2−b2)a2+b2+b2aba2+b2=a3−ab2+2ab2a2+b2=a(a2+b2)a2+b2=a

If $tan x = \frac{b}{a}$ , then the value of $a cos 2 x + b sin 2 x$ is?