Question

Use the graphs to determine which equation could possible be an identity.

Answer 1

A

Explanation:

`(sec t + csc t)/(1 + cot t) = sec t \\\\\ ((1)/(cos t) + (1)/(sin t))/(1 + (cos t)/(sin t)) = (1)/(cos t) \\\\\ (1)/(cos t) + (1)/(sin t) = (1)/(cos t)(1 + (cos t)/(sin t)) \\\\\ (1)/(cos t) + (1)/(sin t) = (1)/(cos t) + (1)/(sin t)`

I hope I helped you.

Answer 2

The correct answer is:

Option: A

A.
`(\sec t+\csc t)/(1+\cot t)=\sec t`

Explanation:

A)

`(\sec t+\csc t)/(1+\cot t)=\sec t\\\\\\i.e.\\\\\\((1)/(\cos t)+(1)/(\sin t))/(1+(\cos t)/(\sin t))=\sec t\\\\\\((\cos t+\sin t)/(\cos t\sin t))/((\sin t+\cos t)/(\sin t))=\sec t\\\\\\(\sin t)/(\sin t\cos t)=\sec t\\\\\\(1)/(\cos t)=\sec t\\\\\\\\\sec t=\sec t`

B)

`(\sec t-\csc t)/(1-\cot t)=\cos t\\\\\\i.e.\\\\\\((1)/(\cos t)-(1)/(\sin t))/(1-(\cos t)/(\sin t))=\cos t\\\\\\((\sin t-\cos t)/(\cos t\sin t))/((\cos t-\sin t)/(\sin t))=\cos t\\\\\\(-\sin t)/(\sin t\cos t)=\cos t\\\\\\(-1)/(\cos t)=\cos t\\\\\\\\\-sec t=\cos t`

This is a false relation.

Since, we know that:

`\sec t=(1)/(\cos t)`

C)

`(\cos t-\sin t)/(1-\cot t)=\cos t`

i.e.

`(\cos t-\sin t)/(1-(\cos t)/(\sin t))=\cos t\\\\\\(\cos t-\sin t)/((\sin t-\cos t)/(\sin t))=\cos t\\\\\\-(1)/(\sin t)=\cos t`

which is again a wrong identity

D)

`(\sin (t+2\pi))/(\cos (\pi+t))=\cos t`

We know that:

`\sin (2\pi+t)=\sin t\\\\and\\\\\cos (\pi+t)=-\cos t`

i.e.

`(\sin t)/(-\cos t)=\cos t`

i.e.

`-\csc t=\cos t`

which is again a wrong identity.