Question

SOLVE THIS EQUATION FAST...​

Answer 1

Question:-

If √3 tan θ = 3 sin θ then prove that sin² θ - cos² θ = 1/3.

Answer:-

Given:-

√3 tan θ = 3 sin θ

$\implies \sf \: \frac{ \sqrt{3} \times \tan \theta }{ \sin \theta } = 3 \\ \\ \\ \implies \sf \: \frac{ \tan \theta}{ \sin \theta } = \frac{3}{ \sqrt{3} }$

Using tan θ = sin θ/cos θ in LHS we get,

$\\ \implies \sf \: \frac{ \frac{ \sin \theta }{ \cos \theta} }{ \sin \theta} = \sqrt{3} \\ \\ \\ \implies \sf \: \frac{ \sin \theta }{ \cos \theta} \times \frac{1}{ \sin \theta} = \sqrt{3} \\ \\ \\ \implies \sf \: \frac{1}{ \cos \theta} = \sqrt{3} \\ \\ \\ \implies \boxed{\sf \cos \theta = \frac{1}{ \sqrt{3} } }$

Now,

We have to prove:-

sin² θ - cos² θ = 1/3

Using sin² θ = 1 - cos² θ we get,

⟹ 1 - cos² θ - cos² θ = 1/3

Putting the value of cos θ we get,

⟹ 1 - (1/√3)² - (1/√3)² = 1/3

⟹ 1 - 1/3 - 1/3 = 1/3

⟹ (3 - 1 - 1)/3 = 1/3

⟹ 1/3 = 1/3

Hence, Proved.

Answer 2

${\large{\bold{\rm{\underline{Question}}}}}$

${\red{\bigstar}}$ If ${\sf{\sqrt{3}}}$ tan θ = 3 sin θ then prove that sin²θ - cos² θ = ${\sf{\dfrac{1}{3}}}$

${\large{\bold{\rm{\underline{Required \; Solution}}}}}$

${\sf{\green{Given \; that}}}$

${\bf{:\rightarrow \sqrt{3} tan \theta = \: 3 \: sin \: \theta}}$

${\sf{\green{Let's \; solve \; it}}}$

${\bf{:\rightarrow \sqrt{3} \: \dfrac{sin \theta}{cos \theta} \: = 3 \: sin \: \theta}}$

${\bf{:\rightarrow \sqrt{3} sin \theta \: 3 \sqrt{3} \times \sqrt{3} \: sin \: \theta \: cos \: \theta}}$

${\bf{:\rightarrow \dfrac{\sqrt{3}sin \theta}{\sqrt{3}sin \theta} \: = \sqrt{3} cos \theta}}$

∴ ${\bf{:\rightarrow cos \theta \: = \dfrac{1}{\sqrt{3}}}}$ ${\tt{\pink{Equation \: 1}}}$

${\sf{\green{Now \: let's \: square \: on \: both \: side \: of \: Equation \: 1}}}$

${\bf{:\rightarrow cos^{2} \theta \: = \: \dfrac{1}{3}}}$ ${\tt{\pink{Equation \: 2}}}$

${\sf{\green{Now \: we \: know \: that}}}$

${\bf{:\rightarrow cos^{2} \theta = \: 1 - sin^{2} \theta}}$

${\bf{:\rightarrow 1 - sin^{2} \theta \: = \: \dfrac{1}{3}}}$ ${\tt{\pink{From \: equation \: 2}}}$

${\bf{:\rightarrow sin^{2} \theta \: = 1 - \dfrac{1}{3}}}$

${\bf{:\rightarrow sin^{2} \theta \: = \dfrac{2}{3}}}$ ${\tt{\pink{Equation \: 3}}}$

${\sf{\green{Now,}}}$

${\bf{:\rightarrow sin^{2} \theta \: - cos^{2} \theta}}$

${\bf{:\rightarrow \dfrac{2}{3} - \dfrac{1}{3}}}$

${\bf{:\rightarrow \dfrac{2-3}{1}}}$

${\bf{:\rightarrow \dfrac{1}{3}}}$ ${\tt{\pink{Henceforth, \: proved}}}$ !