Question

Please someone help me...​

Answer 1

Answer:   see proof below

Step-by-step explanation:

Use the following identities:

$\cot\alpha=\dfrac{1}{\tan\alpha}\\\\\\\cot(\alpha-\beta)=\dfrac{1+\tan\alpha\cdot \tan\beta}{\tan\alpha-\tan\beta}$

Proof  LHS →  RHS

Given:                  $\dfrac{1}{\tan 6A-\tan 2A}-\dfrac{1}{\cot 6A-\cot 2A}$

Cot Identity:        $\dfrac{1}{\tan 6A-\tan 2A}-\dfrac{1}{\dfrac{1}{\tan 6A}-\dfrac{1}{\tan 2A}}$

Simplify:              $\dfrac{1}{\tan 6A-\tan 2A}-\dfrac{1}{\dfrac{1}{\tan 6A}\bigg(\dfrac{\tan 2A}{\tan 2A}\bigg)-\dfrac{1}{\tan 2A}\bigg({\dfrac{\tan 6A}{\tan 6A}\bigg)}}$

                         $= \dfrac{1}{\tan 6A-\tan 2A}-\dfrac{1}{\dfrac{\tan 2A-\tan 6A}{\tan 6A\cdot \tan 2A}}$

                         $= \dfrac{1}{\tan 6A-\tan 2A}-\dfrac{\tan6A\cdot \tan 2A}{\tan 2A-\tan 6A}$

                         $= \dfrac{1}{\tan 6A-\tan 2A}-\dfrac{\tan6A\cdot \tan 2A}{\tan 2A-\tan 6A}\bigg(\dfrac{-1}{-1}\bigg)$

                        $= \dfrac{1}{\tan 6A-\tan 2A}+\dfrac{\tan6A\cdot \tan 2A}{\tan 6A-\tan 2A}$

                        $= \dfrac{1+\tan6A\cdot \tan 2A}{\tan 6A-\tan 2A}$

Sum Difference Identity:    cot(6A - 2A)

Simplify:                               cot 4A

cot 4A = cot 4A   $\checkmark$

Answer 2

Step-by-step explanation:

First factor out the negative sign from the expression and reorder the terms

That's

$\frac{1}{ - (( \tan(2A) - \tan(6A) )} - \frac{1}{ \cot(6A) - \cot(2A) }$

Using trigonometric identities

That's

$\cot(x) = \frac{1}{ \tan(x) }$

Rewrite the expression

That's

$\frac{1}{ - (( \tan(2A) - \tan(6A) )} - \frac{1}{ \frac{1}{ \tan(6A) } } - \frac{1}{ \frac{1}{ \tan(2A) } }$

We have

$- \frac{1}{ \tan(2A) - \tan(6A) } - \frac{1}{ \frac{ \tan(2A) - \tan(6A) }{ \tan(6A) \tan(2A) } }$

Rewrite the second fraction

That's

$- \frac{1}{ \tan(2A) - \tan(6A) } - \frac{ \tan(6A) \tan(2A) }{ \tan(2A) - \tan(6A) }$

Since they have the same denominator we can write the fraction as

$- \frac{1 + \tan(6A) \tan(2A) }{ \tan(2A) - \tan(6A) }$

Using the identity

$\frac{x}{y} = \frac{1}{ \frac{y}{x} }$

Rewrite the expression

We have

$- \frac{1}{ \frac{ \tan(2A) - \tan(6A) }{1 + \tan(6A) \tan(2A) } }$

Using the trigonometric identity

$\frac{ \tan(x) - \tan(y) }{1 + \tan(x) \tan(y) } = \tan(x - y)$

Rewrite the expression

That's

$- \frac{1}{ \tan(2A -6A) }$

Which is

$- \frac{1}{ \tan( - 4A) }$

Using the trigonometric identity

$\frac{1}{ \tan(x) } = \cot(x)$

Rewrite the expression

That's

$- \cot( - 4A)$

Simplify the expression using symmetry of trigonometric functions

That's

$- ( - \cot(4A) )$

Remove the parenthesis

We have the final answer as

$\cot(4A)$

As proven

Hope this helps you