Solve this

Solution:

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