Question

Please someone help me...​

Answer 1

Answer: see proof below

Explanation:

Use the following identities:

`\cot\alpha=(1)/(\tan\alpha)\\\\\\\cot(\alpha-\beta)=(1+\tan\alpha\cdot \tan\beta)/(\tan\alpha-\tan\beta)`

Proof LHS → RHS

Given:
`(1)/(\tan 6A-\tan 2A)-(1)/(\cot 6A-\cot 2A)`

Cot Identity:
`(1)/(\tan 6A-\tan 2A)-(1)/((1)/(\tan 6A)-(1)/(\tan 2A))`

Simplify:
`(1)/(\tan 6A-\tan 2A)-\frac{1}{(1)/(\tan 6A)\bigg((\tan 2A)/(\tan 2A)\bigg)-(1)/(\tan 2A)\bigg({(\tan 6A)/(\tan 6A)\bigg)}}`

`= (1)/(\tan 6A-\tan 2A)-(1)/((\tan 2A-\tan 6A)/(\tan 6A\cdot \tan 2A))`

`= (1)/(\tan 6A-\tan 2A)-(\tan6A\cdot \tan 2A)/(\tan 2A-\tan 6A)`

`= (1)/(\tan 6A-\tan 2A)-(\tan6A\cdot \tan 2A)/(\tan 2A-\tan 6A)\bigg((-1)/(-1)\bigg)`

`= (1)/(\tan 6A-\tan 2A)+(\tan6A\cdot \tan 2A)/(\tan 6A-\tan 2A)`

`= (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

Explanation:

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

That's

`(1)/( - (( \tan(2A) - \tan(6A) )) - (1)/( \cot(6A) - \cot(2A) )`

Using trigonometric identities

That's

`\cot(x) = (1)/( \tan(x) )`

Rewrite the expression

That's

`(1)/( - (( \tan(2A) - \tan(6A) )) - (1)/( (1)/( \tan(6A) ) ) - (1)/( (1)/( \tan(2A) ) )`

We have

`- (1)/( \tan(2A) - \tan(6A) ) - (1)/( ( \tan(2A) - \tan(6A) )/( \tan(6A) \tan(2A) ) )`

Rewrite the second fraction

That's

`- (1)/( \tan(2A) - \tan(6A) ) - ( \tan(6A) \tan(2A) )/( \tan(2A) - \tan(6A) )`

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

`- (1 + \tan(6A) \tan(2A) )/( \tan(2A) - \tan(6A) )`

Using the identity

`(x)/(y) = (1)/( (y)/(x) )`

Rewrite the expression

We have

`- (1)/( ( \tan(2A) - \tan(6A) )/(1 + \tan(6A) \tan(2A) ) )`

Using the trigonometric identity

`( \tan(x) - \tan(y) )/(1 + \tan(x) \tan(y) ) = \tan(x - y)`

Rewrite the expression

That's

`- (1)/( \tan(2A -6A) )`

Which is

`- (1)/( \tan( - 4A) )`

Using the trigonometric identity

`(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