`1/(sin2A)- 1/(tan2A)`

Simplify using the formula for double angles where

`sin 2A = 2sinAcosA `

Remember that `tan=sin/cos` `therefore 1/(tan2A) = 1/((sin2A)/(cos2A))`

`therefore1/(tan2A) = (cos2A)/(sin2A)`

From the formula for double angles(use the most relevant for cos ie `1-2sin^2A` )

`(cos2A)/(sin2A)= (1-2sin^2A)/(2sinAcosA)`

Now put thses results into the original expression:

`1/(2sinAcosA) - (1-2sin^2A)/(2sinAcosA)`

The denominator is the same

`therefore = (1-1-2sin^2A)/(2sinAcosA)`

`therefore = (-2sin^2A)/(2sinAcosA)`

Simplify as the 2s cross cancel and one of the sinAs

`therefore = -sinA/cosA`

We know that `sinA/cosA=tan A`

`therefore = -tanA`

**The answer is tan A**

Given expression is `1/(sin2A)-1/(tan2A)`

Now we know that `tan2A=(sin2A)/(cos2A)`

So we can write `1/(sin2A)-1/(tan2A)=1/(sin2A)-1/((sin2A)/(cos2A))`

`=1/(sin2A)-(cos2A)/(sin2A)`

`=(1-cos2A)/(sin2A)`

( `cos2A=1-2(sin^2A)` )

`=(1-(1-2sin^2A))/(sin2A)`

`=(1-1+2sin^2A)/(sin2A)`

As `sin2A=2(sinA.cosA)`

` =(2sin^2A)/(2sinA.cosA) ` `=(sinA)/(cosA)` `=tanA` . Answer.

`1/(sin(2A))-1/(tan(2A))` (i)

We know `tan(2A)=(sin(2A))/(cos(2A))`

`1/(tan(2A))=cos(2A)/sin(2A)`

`` So (i) ,reduces to

`1/sin(2A)-cos(2A)/sin(2A)=(1-cos(2A))/sin(2A)` (ii)

we also know , `sin(2A)=2sin(A)cos(A) `

`and cos(2A)=1-2sin^2(A)`

So substitute in (ii) ,it will reduce to

`(2sin^2(A))/(2sin(A)cos(A))`

`=sin(A)/cos(A)`

`=tanA`