Question

1 / tan²Ꮎ write in terms of sin Ꮎ .​

Answer 1

`\huge\begin{array}{ccc}(1)/(\tan^2\theta)=(1-\sin^2\theta)/(\sin^2\theta)=(1)/(\sin^2\theta)-1\end{array}`

Trigonometry.

We know:

`\sin^2x+\cos^2x=1\\\\\text{tan}x=(\sin x)/(\cos x)`

SOLUTION:

We have:

`(1)/(\tan^2\theta)`

write it in therms of
`\sin\theta`.

`\tan\theta=(\sin\theta)/(\cos\theta)\Rightarrow\text{tan}^2\theta=\left((\sin\theta)/(\cos\theta)\right)^2=(\sin^2\theta)/(\cos^2\theta)`

`\sin^2\theta+\cos^2\theta=1\to\cos^2\theta=1-\sin^2\theta`

therefore:

`\tan^2\theta=(\sin^2\theta)/(\cos^2\theta)=(\sin^2\theta)/(1-\sin^2\theta)`

Finally:

`(1)/(\tan^2\theta)=(1)/((\sin^2\theta)/(1-\sin^2\theta))=(1-\sin^2\theta)/(\sin^2\theta)=(1)/(\sin^2\theta)-(\sin^2\theta)/(\sin^2\theta)=(1)/(\sin^2\theta)-1`