If $\tan \theta=\frac{a}{b}$, prove that $a \sin 2 \theta+b \cos 2 \theta=b$
Given: $\theta=\frac{a}{b}$
To Prove: a sin 2θ + b cos 2θ = b
Given: $\theta=\frac{a}{b}$
We know that
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}=\frac{\mathrm{a}}{\mathrm{b}}$
By Pythagoras Theorem,
(Perpendicular) $^{2}+(\text { Base })^{2}=(\text { Hypotenuse })^{2}$
$\Rightarrow(a)^{2}+(b)^{2}=(H)^{2}$
$\Rightarrow a^{2}+b^{2}=(H)^{2}$
$\Rightarrow H=\sqrt{a^{2}+b^{2}}$
So,
$\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{\mathrm{a}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}$
$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}$
Taking LHS,
= a sin 2θ + b cos 2θ
We know that,
$\sin 2 \theta=2 \sin \theta \cos \theta$
and $\cos 2 \theta=1-2 \sin ^{2} \theta$
$=a(2 \sin \theta \cos \theta)+b\left(1-2 \sin ^{2} \theta\right)$
Putting the values of sinθ and cosθ, we get
$=a \times 2 \times \frac{a}{\sqrt{a^{2}+b^{2}}} \times \frac{b}{\sqrt{a^{2}+b^{2}}}+b\left[1-2 \times\left(\frac{a}{\sqrt{a^{2}+b^{2}}}\right)^{2}\right]$
$=\frac{2 a^{2} b}{a^{2}+b^{2}}+b\left[1-2 \times \frac{a^{2}}{a^{2}+b^{2}}\right]$
$=\frac{2 a^{2} b}{a^{2}+b^{2}}+b-\frac{2 a^{2} b}{a^{2}+b^{2}}$
= b
= RHS
∴ LHS = RHS
Hence Proved
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