While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods. 

The correct answer to how to best begin this verification is "Multiply the numerator and denominator of a fraction by the conjugate of each other." This technique will be effective because initially, the denominators are pretty different, but furthermore, it will help turn  into , which is a Pythagorean identity that can be replaced by . You can see this demonstrated in full in the full verification below:

Let's also explore why the incorrect answers are incorrect. 

Firstly, while getting everything in terms of sines and cosines is a popular strategy in verifying trig identities, it won't help us here because everything is already in terms of sines and cosines. Secondly, in order to substitute a Pythagorean Identity, we usually want to see either , , or . None of these are present in our expression from the get-go, so this is a non-starter (though you may note that we used one of these in the middle of the proof. Finally, there are not any terms present that we can factor, and there are not any terms present that we can reduce. 

While it is important to understand that there is more than one correct way to verify an identity, certain procedures best apply to certain expressions and will allow you to verify an identity more simply than other methods. 

The correct answer to how to best begin this verification is factor the expression. No matter which side you choose to begin with, you have the ability to factor since  goes into both  and  and since  goes into both  and . That said, be sure to perform work on only one side of the equation; once you've selected one, do not be tempted to modify the other! The verification of this identity is demonstrated below:

Let's also explore why the incorrect answers are incorrect. 

Firstly, while getting everything in terms of sines and cosines is a popular strategy in verifying trig identities, it will actually introduce more fractions. We can also ask ourselves "Why is getting everything in terms of sines and cosines so helpful?" It's helpful because it consolidates expressions that may have 3 to 6 different trig functions down to having only 1 or 2 trig functions (sine, cosine, both). However, in our problem, to begin, we only have two different trig functions (tangent and secant). There are ways to solve this problem such that you could begin with converting to sine and cosine as your first step, but, it would be a messier and less concise way to verify this trig identity.

The next incorrect answer offers to "combine like terms." However, there are no like terms present to combine.

The next option offers to "Multiply the numerator and denominator of a fraction by the conjugate of each other." This is not relevant here, because we do not have any fractions present. Likewise, the option to "reduce the fraction" is not helpful here. 

Notice that there is no answer choice offering to "Substitute a Pythagorean Identity," and this is intentional, because you can absolutely verify this identity by substituting the identity . A brief proof is below, and is a good example of how there can be more than one correct way to verify a trig identity!

Cosine and sine are not reciprocal functions.  

 and 

We begin with the left hand side of the equation and utilize basic trigonometric identities, beginning with converting the inverse functions to their corresponding base functions:

Next we rewrite the fractional division in order to simplify the equation:

In fractional division we multiply by the reciprocal as follows:

If we reduce the fraction using basic identities we see that the equivalence is proven:

The true identity is  because cosine is an even function.

The definition of tangent is sine divided by cosine.

Using these basic identities:

we find the original expression to be

which simplifies to

.

Further simplifying:

The cosines cancel, giving us

Write the Pythagorean identity.

Substract  from both sides.

The other answers are incorrect.

The correct answer is . Begin by substituting , , and . This gives us:

.

To solve this problem, use the identities , , , and . Then we get