Question

How to rewrite cot(squared)-tan(squared) in terms of sin?

Answer 1

First things first, we know that

`cotx=1/tanx`

Hence,

`cot^2x-tan^2x=1/tan^2x-tan^2x`

`=(1-tan^4x)/tan^2`

Since we have a difference of two squares, we can apply the formula:

`color(red)(1-tan^4x=(1-tan^2x)(1+tan^2x)`

`:.cot^2x-tan^2x=((1-tan^2x)(1+tan^2x))/tan^2x`

`diamond 1-tan^2x=1-sin^2x/cos^2x=(cos^2x-sin^2x)/cos^2x`

`diamond 1+tan^2x=1+sin^2x/cos^2x=(cos^2x+sin^2x)/cos^2x=1/cos^2x`

`:.cot^2x-tan^2x=(cos^2x-sin^2x)/(sin^2x/cos^2xcos^4x)=(cos^2x-sin^2x)/(sin^2xcos^2x)`

Since `cos^2x=1-sin^2x`:

`:. cot^2x-tan^2x=(1-2sin^2x)/(sin^2x(1-sin^2x))`

`=1/sin^2x-1/(1-sin^2x)`

Answer 2

Answer for the question :
Rewrite : `cot^2x-tan^2x` in terms of `sinx`

`(i)cot^2x-tan^2x=1/sin^2x-1/(1-sin^2x) or`

`(ii)cot^2x-tan^2x=(1-2sin^2x)/(sin^2x(1-sin^2x))`

Explanation:

We know that,

`color(red)((1)csc^2theta-cot^2theta=1) and color(blue)((2)sec^2theta-tan^2theta=1)`

Using `(1) and (2)`

`cot^2x-tan^2x=color(red)((csc^2x-1))-color(blue)((sec^2x-1))`

`color(white)(cot^2x-tan^2x)=csc^2x-sec^2x`

`color(white)(cot^2x-tan^2x)=1/sin^2x-1/cos^2x`

`color(white)(cot^2x-tan^2x)=1/sin^2x-1/(1-sin^2x)`

#color(white)(cot^2x-tan^2x)=((1-sin^2x)-sin^2x)/(sin^2x(1- sin^2x))#

`color(white)(cot^2x-tan^2x)=(1-2sin^2x)/(sin^2x(1-sin^2x))`