Question

Hello, I need some help resolving this problem of Trigonometric Identities. Use the reciprocal identities to resolve it SinA+cosA*cotA= cscA

Answer 1

Please see steps below

Explanation:

Start by writing all trig functions in the equation in terms of their simplest forms using the two basic trig functions:
`sin(\alpha) \,\,and\,\,cos(\alpha)`:

`sin(\alpha)+ cos(\alpha)\,(cos(\alpha))/(sin(\alpha)) = (1)/(sin(\alpha))`

Now work on the left side (which is the most complicated one), trying to simplify it using the properties for adding fractions with different denominators:

`sin(\alpha)+ cos(\alpha)\,(cos(\alpha))/(sin(\alpha))=sin(\alpha)+(cos^2(\alpha))/(sin(\alpha)) =(sin^2(\alpha))/(sin(\alpha)) +(cos^2(\alpha))/(sin(\alpha))=(sin^2(\alpha)+cos^2(\alpha))/(sin(\alpha))=(1)/(sin(\alpha))`

where in the last step we have used that the Pythagorean identity for:

`sin^2(\alpha)+cos^2{\alpha)=1`

Notice that we arrived at the expression:
`(1)/(sin(\alpha))`, which is exactly what appears on the other side of the initial equation/identity we needed to prove, so the prove has been completed.