Question

please solve this question



prove it​

Answer 1

$\boxed{\textbf{\large{To prove :}}}$

$\frac{ \sin(2x) }{1 - \cos(2x) } = \cot(x)$

$\boxed{\textbf{\large{proof :}}}$

here remember that,

➡ $(\sin(2x) = 2 \sin(x) . \cos(x))$

And

➡ $(1 - \cos(2x) = 2 { \sin }^{2} (x))$

Therefor ,

LHS =

$=\frac{ \sin(2x) }{1 - \cos(2x) }$

$= \frac{2 \sin(x) \cos(x) }{2 { \sin}^{2} x}$

$= \frac{ \cos(x) }{ \sin(x) }$

$= \cot(x)$

= RHS

hence proved .

_________________________________

Answer 2

Step-by-step explanation:

From trigonometric ratios of compound angles we have known that

➡➡Sin(A+B)=sinAcosB+cosAsinB

If here we take the same angle A it will be

➡➡ Sin(A+A)=sinAcosA+cosAsinA

➡➡Or, sin2A= sinA.2cosA ........(i)

And

From Another formula

Cos(A+B)=cosAcosB - sinAsInB

Or, cos(A+A)=cosAcosA - sinAsinA

$or \: cos2a = {cos}^{2} a - {sin}^{2} a \\ \\ = > \cos(2a) = 1 - {sin}^{2} a - {sin}^{2} a \\ \\ = > \cos(2a) = 1 - 2 {sin}^{2} a \: .........(2)$

To proof that :

LHS:---

$\frac{sin2x}{1 - cos2x} \\ \\ = \frac{2 \sin(x) \cos(x) }{1 - 1 + 2 {sin}^{2}x } \\ \\ = > \frac{ \cos(x) }{ \sin(x) } \\ \\ = > cotx(rhs)$

hope it helps✨✨✨✨