Verify that each equation is an identity.

sin θ + cos θ = sin θ/(1 - cot θ) + cos θ/(1 - tan θ)

Video transcript

Hello, everyone. We are asked if the given equation is an identity, we are given the cosine of X divided by one plus the tangent of X plus the sine of X divided by one plus the cotangent of X equals one divided by the sine of X plus the cosine of X. Our answer choices are a yes B no. All right. I'm going to rewrite my original equation. So I have the cosine of X divided by one plus the tangent of X plus the sine of X divided by one plus the cotangent of X. And actually, I'm gonna stop right there with the left hand side because to make this an identity, I need to make the left hand side match the right hand side. So let's work with the left hand side. I'm going to take my tangent and cotangent and rewrite them in terms of sign and cosine. So this would be rewritten as the cosine of X divided by one plus the sine of X divided by the cosine of X plus the sign of X divided by one plus the cosine of X divided by the sine of X. Now that everything is in terms of sign and cosign. I'm going to make a common denominator in my denominators. So the name reader of the cosine of X stays put in the denominator. I'm going to rewrite one as cosine X divided by cosine X plus the sine of X divided by the cosine of X in my second fraction. Our numerator of sine X stays put, my one will be rewritten as the sign of X divided by the sign of X to make a common denominator plus the cosine of X divided by the sine of X. I am going to add my denominators. So I'll have the cosine of X divided by the cosine of X plus the sine of X divided by the cosine of X plus the sine of X divided by the sine of X plus the cosine of X all divided by the sine of X. So from there, I'm gonna go back up to the top, just have a little more space. I'm going to rewrite my complex fractions as division problems. So in parentheses of the cosine of X divided by one divided by the cosine of X plus the sine of X divided by the cosine of X plus. The next set of parentheses will be the sine of X divided by one divided by the sine of X plus the cosine of X divided by the sign of X closed parentheses from there. I'm changing my division into multiplying by the reciprocal. So I'll have the cosine of X divided by one multiplied by the cosine of X divided by the cosine of X plus the sine of X closed parentheses plus next set of parentheses, the sine of X divided by one multiplied by the sine of X divided by the sine of X plus the cosine of X close parentheses. Multiplying this out, I get cosine X times cosine X which when you multiply cosine X by itself, you'll get cosine squared X divided by the cosine of X plus the sine of X plus sine of X multiplied by sine of X will be sine squared X divided by the sine of X plus the cosine of X. Now, despite the fact that cosine and sine are in opposite orders in our denominator. In addition, the order doesn't matter. So these do have a common denominator already and I'm going to rewrite it as sine of X plus cosine of X is my denominator. In the numerator, I have cosine squared X plus sine squared X. And I recall that there is a Pythagorean identity that states that the cosine squared X plus the sine squared of X equals one. So I could replace my numerator with the value one and that'll still be divided by the sine of X plus the cosine of X and now if I check this matches my right hand side, so since this is the right hand side and they match exactly this is answer choice. A yes, it is an identity. Have a nice day.