Question

2.)
A.) 1/ sin 0
B.) 1/ cot 0
C.) 1/ cos 0
D.) 1/ csc 0
E.) 1/ sec 0

Answer 1

B)
`(1)/(cot\theta)`.

Explanation:

We have to find out the reciprocal of
`tan\theta`.

We have drawn a triangle for your reference.

In which
`\angle C=\theta`

AB = opposite side

BC = adjacent side

CA = hypotenuse

Since we know that the
`tan\theta` is equal to opposite side upon adjacent side.

`tan\theta=(opposite\ side)/(adjacent\ side)=(AB)/(BC)`

Or
`tan\theta=(sin\theta)/(cos\theta) \ \ \ \ equation\ 1`

Where as the
`cot\theta` is equal to adjacent side upon opposite side.

Therefore,

`cot\theta=(adjacent\ side)/(opposite\ side)=(BC)/(AB)`

Or
`cot\theta=(cos\theta)/(sin\theta) \ \ \ \ equation \ 2`

From equation 1 and equation 2 we can say that;

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

Hence
`tan\theta=(1)/(cot\theta)`.