Question

How do you simplify the expression `cott(tant+cott)`?

Answer 1

`csc^2t`

Explanation:

We can begin by using the `color(blue)"trigonometric identities"`

`color(orange)"Reminder"`

`color(red)(bar(ul(|color(white)(a/a)color(black)(tant=(sint)/(cost)" and " cott=(cost)/(sint))color(white)(a/a)|)))`

The expression may now be written as.

`(cost)/(sint)((sint)/(cost)+(cost)/(sint))`

and distributing the bracket gives.

`1+(cos^2t)/(sin^2t)=(sin^2t)/(sin^2t)+(cos^2t)/(sin^2t)`

both fractions have a common denominator so adding gives.

`(sin^2t+cos^2t)/(sin^2t)`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(sin^2t+cos^2t=1)color(white)(a/a)|)))`

and `color(red)(bar(ul(|color(white)(a/a)color(black)(csct=1/(sint))color(white)(a/a)|)))`

`rArr(sin^t+cos^2t)/(sin^2t)=1/(sin^2t)=csc^2t`

Thus `cott(tant+cott)" simplifies to" csc^2t`