Question

First question, thanks. I believe there should be 3 answers

Answer 1

Given: The following functions

`A)cos^2\theta=sin^2\theta-1`
`B)sin\theta=(1)/(csc\theta)`
`\begin{gathered} C)sec\theta=(1)/(cot\theta) \\ D)cot\theta=(cos\theta)/(sin\theta) \\ E)1+cot^2\theta=csc^2\theta \end{gathered}`

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

`\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}`

B

`\begin{gathered} sin\theta=(1)/(csc\theta) \\ Note\text{ that} \\ csc\theta=(1)/(sin\theta) \\ sin\theta* csc\theta=1 \\ sin\theta=(1)/(csc\theta) \\ Therefore \\ sin\theta=(1)/(csc\theta),is\text{ an identities} \end{gathered}`

C

`\begin{gathered} sec\theta=(1)/(cot\theta) \\ note\text{ that} \\ cot\theta=(1)/(tan\theta) \\ tan\theta cot\theta=1 \\ tan\theta=(1)/(cot\theta) \\ Therefore, \\ sec\theta\\e(1)/(cot\theta),NOT\text{ IDENTITY} \end{gathered}`

D

`\begin{gathered} cot\theta=(cos\theta)/(sin\theta) \\ Note\text{ that} \\ cot\theta=(1)/(tan\theta) \\ cot\theta=1/ tan\theta \\ tan\theta=(sin\theta)/(cos\theta) \\ So, \\ cot\theta=1/(sin\theta)/(cos\theta) \\ cot\theta=1*(cos\theta)/(sin\theta) \\ cot\theta=(cos\theta)/(sin\theta) \\ Therefore \\ cot\theta=(cos\theta)/(sin\theta),is\text{ an Identity} \end{gathered}`

E

`\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=(1)/(sin^2\theta) \\ cot^2\theta=(cos^2\theta)/(sin^2\theta) \\ So, \\ (1)/(sin^2\theta)-(cos^2\theta)/(sin^2\theta) \\ (1-cos^2\theta)/(sin^2\theta) \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ (1-cos^2\theta)/(sin^2\theta)=(sin^2\theta)/(sin^2\theta)=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}`

Hence, the following are identities

`\begin{gathered} B)sin\theta=(1)/(csc\theta) \\ D)cot\theta=(cos\theta)/(sin\theta) \\ E)1+cot^2\theta=csc^2\theta \end{gathered}`

The marked are the trigonometric identities