Algebra 3 Quiz: Fundamental Trig Identities
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Fundamental Trig IdentitiesQuestion 1 of 20

Which expression is equivalent to sin⁡θ1+cos⁡θ\frac{\sin \theta}{1+\cos \theta} for all values of the variable for which the expression is defined?

1−cos⁡θsin⁡θ\frac{1-\cos \theta}{\sin \theta}
1+cos⁡θsin⁡θ\frac{1+\cos \theta}{\sin \theta}
sin⁡θ1−cos⁡θ\frac{\sin \theta}{1-\cos \theta}
1−sin⁡θcos⁡θ\frac{1-\sin \theta}{\cos \theta}
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Algebra 3 Quiz

Algebra 3 Quiz: Fundamental Trig Identities

Practice Fundamental Trig Identities in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Fundamental Trig Identities, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which expression is equivalent to sin⁡θ1+cos⁡θ\frac{\sin \theta}{1+\cos \theta} for all values of the variable for which the expression is defined?

  1. 1−cos⁡θsin⁡θ\frac{1-\cos \theta}{\sin \theta} (correct answer)
  2. 1+cos⁡θsin⁡θ\frac{1+\cos \theta}{\sin \theta}
  3. sin⁡θ1−cos⁡θ\frac{\sin \theta}{1-\cos \theta}
  4. 1−sin⁡θcos⁡θ\frac{1-\sin \theta}{\cos \theta}
Explanation: Whenever you see a single trig fraction, try multiplying by a conjugate. Since the denominator is 1+cos⁡θ1+\cos\theta, multiply the numerator and denominator by 1−cos⁡θ1-\cos\theta: sin⁡θ1+cos⁡θ⋅1−cos⁡θ1−cos⁡θ=sin⁡θ(1−cos⁡θ)1−cos⁡2θ=sin⁡θ(1−cos⁡θ)sin⁡2θ=1−cos⁡θsin⁡θ.\frac{\sin\theta}{1+\cos\theta}\cdot \frac{1-\cos\theta}{1-\cos\theta} = \frac{\sin\theta(1-\cos\theta)}{1-\cos^2\theta} = \frac{\sin\theta(1-\cos\theta)}{\sin^2\theta} = \frac{1-\cos\theta}{\sin\theta}. The key step uses the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1. So for values where these expressions are defined, 1−cos⁡θsin⁡θ\frac{1-\cos\theta}{\sin\theta} is the equivalent form. The other choices miss this idea. 1+cos⁡θsin⁡θ\frac{1+\cos\theta}{\sin\theta} is what you get if you multiply by 1+cos⁡θ1+\cos\theta instead, and it is actually paired with sin⁡θ1−cos⁡θ\frac{\sin\theta}{1-\cos\theta}; those two are equivalent to each other, not to the original. 1−sin⁡θcos⁡θ\frac{1-\sin\theta}{\cos\theta} comes from incorrectly replacing parts of sin⁡2+cos⁡2=1\sin^2+\cos^2=1, but 1−sin⁡θ1-\sin\theta is not a useful identity here. Study tip: for a denominator like 1+cos⁡θ1+\cos\theta, multiply by its conjugate 1−cos⁡θ1-\cos\theta. This turns the denominator into sin⁡2θ\sin^2\theta and usually lets you cancel to a simpler form.

Question 2

Simplify tan⁡2θsec⁡2θ+cot⁡2θcsc⁡2θ\frac{\tan^2 \theta}{\sec^2 \theta}+\frac{\cot^2 \theta}{\csc^2 \theta} for all values of the variable for which the expression is defined.

  1. sin⁡2θ\sin^2 \theta
  2. 11 (correct answer)
  3. cos⁡2θ\cos^2 \theta
  4. tan⁡2θ\tan^2 \theta
Explanation: When you see nested trig fractions, your first move should be to rewrite everything in terms of sine and cosine; this usually exposes a hidden Pythagorean identity. Start with tan⁡2θ=sin⁡2θcos⁡2θ\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta} and sec⁡2θ=1cos⁡2θ\sec^2\theta=\frac{1}{\cos^2\theta}, so tan⁡2θsec⁡2θ=sin⁡2θ/cos⁡2θ1/cos⁡2θ=sin⁡2θ.\frac{\tan^2\theta}{\sec^2\theta} =\frac{\sin^2\theta/\cos^2\theta}{1/\cos^2\theta} =\sin^2\theta. Similarly, cot⁡2θ=cos⁡2θsin⁡2θ\cot^2\theta=\frac{\cos^2\theta}{\sin^2\theta} and csc⁡2θ=1sin⁡2θ\csc^2\theta=\frac{1}{\sin^2\theta}, so cot⁡2θcsc⁡2θ=cos⁡2θ/sin⁡2θ1/sin⁡2θ=cos⁡2θ.\frac{\cot^2\theta}{\csc^2\theta} =\frac{\cos^2\theta/\sin^2\theta}{1/\sin^2\theta} =\cos^2\theta. Adding gives sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1. The choice sin⁡2θ\sin^2\theta is only the first term, not the whole sum. The choice cos⁡2θ\cos^2\theta is only the second term. The choice tan⁡2θ\tan^2\theta might tempt you if you combine the fractions incorrectly or think the expression simplifies to a tangent identity, but after rewriting in sine and cosine, no tangent remains in the simplified result. Your takeaway: when simplifying a trig expression, convert everything to sin⁡θ\sin\theta and cos⁡θ\cos\theta first, then look for sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1. This approach turns a complicated-looking fraction into a familiar identity almost every time.

Question 3

Which expression is NOT equivalent to sin⁡2x\sin^2 x for all values of xx for which it is defined?

  1. 1−cos⁡2x1-\cos^2 x
  2. tan⁡2xsec⁡2x\frac{\tan^2 x}{\sec^2 x}
  3. 1sec⁡2x\frac{1}{\sec^2 x} (correct answer)
  4. 1csc⁡2x\frac{1}{\csc^2 x}
Explanation: When you see a question asking whether an expression is equivalent to sin⁡2x\sin^2 x, your first move should be to rewrite everything in terms of sine and cosine. That instantly makes most equivalences clear. The expression 1−cos⁡2x1-\cos^2 x is just a rearrangement of the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, so it is exactly sin⁡2x\sin^2 x. For tan⁡2xsec⁡2x\frac{\tan^2 x}{\sec^2 x}, rewrite it as sin⁡2x/cos⁡2x1/cos⁡2x\frac{\sin^2 x/\cos^2 x}{1/\cos^2 x}, which simplifies to sin⁡2x\sin^2 x. Similarly, 1csc⁡2x\frac{1}{\csc^2 x} is sin⁡2x\sin^2 x, because csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}, so its reciprocal squared is sin⁡2x\sin^2 x. The trap is 1sec⁡2x\frac{1}{\sec^2 x}. Since sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}, this expression equals cos⁡2x\cos^2 x, not sin⁡2x\sin^2 x. Many students confuse secant with cosecant: secant pairs with cosine, while cosecant pairs with sine. That confusion makes this choice look like it should match, but it is the one expression that is not equivalent. A strong study habit for trig equivalence questions: always convert tan⁡\tan, cot⁡\cot, sec⁡\sec, and csc⁡\csc into sin⁡\sin and cos⁡\cos before simplifying. Memorize the reciprocal pairs — secant is cosine's reciprocal, cosecant is sine's — and the Pythagorean identity. Those two tools will carry you through most of these problems.

Question 4

Simplify csc⁡xcot⁡x+tan⁡x\frac{\csc x}{\cot x+\tan x} for all values of xx for which the expression is defined.

  1. sin⁡x\sin x
  2. cos⁡x\cos x (correct answer)
  3. sec⁡x\sec x
  4. csc⁡x\csc x
Explanation: Whenever you see csc⁡x\csc x, cot⁡x\cot x, and tan⁡x\tan x together, rewrite everything in terms of sine and cosine. Here, csc⁡x=1sin⁡x,cot⁡x=cos⁡xsin⁡x,tan⁡x=sin⁡xcos⁡x.\csc x=\frac1{\sin x},\quad \cot x=\frac{\cos x}{\sin x},\quad \tan x=\frac{\sin x}{\cos x}. So the denominator becomes cot⁡x+tan⁡x=cos⁡xsin⁡x+sin⁡xcos⁡x=cos⁡2x+sin⁡2xsin⁡xcos⁡x=1sin⁡xcos⁡x.\cot x+\tan x =\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x} =\frac{\cos^2 x+\sin^2 x}{\sin x\cos x} =\frac1{\sin x\cos x}. Then the original expression is 1sin⁡x1sin⁡xcos⁡x=1sin⁡x⋅sin⁡xcos⁡x=cos⁡x.\frac{\frac1{\sin x}}{\frac1{\sin x\cos x}} =\frac1{\sin x}\cdot \sin x\cos x =\cos x. So cos⁡x\cos x is the simplified form. The wrong choices come from reciprocal mistakes. sin⁡x\sin x would result from confusing csc⁡x\csc x with sin⁡x\sin x or flipping the division incorrectly. sec⁡x\sec x is the reciprocal of the correct answer; it appears if you forget that dividing by 1sin⁡xcos⁡x\frac1{\sin x\cos x} means multiplying by sin⁡xcos⁡x\sin x\cos x. csc⁡x\csc x just repeats the numerator and ignores that the denominator simplifies to csc⁡xsec⁡x\csc x \sec x, not to 1. Study tip: for expressions with reciprocal trig functions, convert to sine and cosine first, and watch for sin⁡2x+cos⁡2x=1\sin^2 x+\cos^2 x=1. That identity is the key to collapsing the denominator.

Question 5

Simplify (sin⁡x+cos⁡x)2+(sin⁡x−cos⁡x)2(\sin x+\cos x)^2+(\sin x-\cos x)^2.

  1. 2sin⁡xcos⁡x2\sin x\cos x
  2. 00
  3. 2sin⁡2x2\sin^2 x
  4. 22 (correct answer)
Explanation: Whenever you see a squared binomial involving sine and cosine, expand it rather than trying to simplify in your head. This question tests binomial expansion and the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x+\cos^2 x=1. Expand each square: (sin⁡x+cos⁡x)2=sin⁡2x+2sin⁡xcos⁡x+cos⁡2x(\sin x+\cos x)^2=\sin^2 x+2\sin x\cos x+\cos^2 x (sin⁡x−cos⁡x)2=sin⁡2x−2sin⁡xcos⁡x+cos⁡2x(\sin x-\cos x)^2=\sin^2 x-2\sin x\cos x+\cos^2 x Adding them, the middle terms 2sin⁡xcos⁡x2\sin x\cos x and −2sin⁡xcos⁡x-2\sin x\cos x cancel, leaving 2sin⁡2x+2cos⁡2x=2(sin⁡2x+cos⁡2x)=2(1)=2.2\sin^2 x+2\cos^2 x=2(\sin^2 x+\cos^2 x)=2(1)=2. The choice 2sin⁡xcos⁡x2\sin x\cos x is the result of focusing on one cross term and ignoring the cancellation, or of subtracting the two expansions instead of adding. The choice 00 comes from thinking the cross terms cancel completely and forgetting that the squared sine and cosine terms remain. The choice 2sin⁡2x2\sin^2 x is the classic error of distributing the square incorrectly, treating (sin⁡x+cos⁡x)2(\sin x+\cos x)^2 as sin⁡2x+cos⁡2x\sin^2 x+\cos^2 x and (sin⁡x−cos⁡x)2(\sin x-\cos x)^2 as sin⁡2x−cos⁡2x\sin^2 x-\cos^2 x, then adding to get 2sin⁡2x2\sin^2 x. A quick check: plug in x=0x=0. The original gives (0+1)2+(0−1)2=2(0+1)^2+(0-1)^2=2, which confirms the answer. On the exam, expand squared trig binomials fully, let the middle terms cancel, and use sin⁡2x+cos⁡2x=1\sin^2 x+\cos^2 x=1.

Question 6

Simplify 1+cot⁡2x1+tan⁡2x\frac{1+\cot^2 x}{1+\tan^2 x} for all values of xx for which the expression is defined.

  1. tan⁡2x\tan^2 x
  2. csc⁡2x\csc^2 x
  3. sec⁡2x\sec^2 x
  4. cot⁡2x\cot^2 x (correct answer)
Explanation: Seeing 1+cot⁡2x1+\cot^2 x and 1+tan⁡2x1+\tan^2 x should immediately trigger the Pythagorean identities: 1+cot⁡2x=csc⁡2x1+\cot^2 x=\csc^2 x and 1+tan⁡2x=sec⁡2x1+\tan^2 x=\sec^2 x. These identities hold for every xx where the original expression is defined, meaning where both sin⁡x\sin x and cos⁡x\cos x are nonzero. Replacing the numerator and denominator gives csc⁡2xsec⁡2x=1sin⁡2x1cos⁡2x=cos⁡2xsin⁡2x=cot⁡2x.\frac{\csc^2 x}{\sec^2 x} = \frac{\frac{1}{\sin^2 x}}{\frac{1}{\cos^2 x}} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x. So the simplified form is cot⁡2x\cot^2 x. The wrong choices each come from a specific misstep. Choosing tan⁡2x\tan^2 x is the reciprocal of the correct answer; this happens if you invert the fraction instead of simplifying it. Choosing csc⁡2x\csc^2 x treats only the numerator as the answer, ignoring the denominator sec⁡2x\sec^2 x. Choosing sec⁡2x\sec^2 x focuses only on the denominator, or mistakes the denominator alone for the whole simplified expression. For this exam, whenever you see 1+tan⁡2x1+\tan^2 x or 1+cot⁡2x1+\cot^2 x, rewrite them as sec⁡2x\sec^2 x and csc⁡2x\csc^2 x before doing anything else. Also remember cot⁡x=cos⁡xsin⁡x\cot x = \frac{\cos x}{\sin x}, not the reciprocal of tan⁡x\tan x in a way that would lead you to invert the final ratio. That one substitution will make this type of problem straightforward.

Question 7

If sin⁡θ+cos⁡θ=12\sin \theta+\cos \theta=\frac{1}{2}, what is sin⁡θcos⁡θ\sin \theta\cos \theta?

  1. −38-\frac{3}{8} (correct answer)
  2. 38\frac{3}{8}
  3. −34-\frac{3}{4}
  4. 14\frac{1}{4}
Explanation: When you see a question involving both sin⁡θ+cos⁡θ\sin\theta+\cos\theta and sin⁡θcos⁡θ\sin\theta\cos\theta, your first thought should be to square the given equation. That connects the sum to the product through the Pythagorean identity. Start with sin⁡θ+cos⁡θ=12\sin\theta+\cos\theta=\frac12. Square both sides: (sin⁡θ+cos⁡θ)2=14(\sin\theta+\cos\theta)^2=\frac14 Expand: sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=14\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta=\frac14 Since sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, this becomes: 1+2sin⁡θcos⁡θ=141+2\sin\theta\cos\theta=\frac14 Subtract 1 from both sides: 2sin⁡θcos⁡θ=−342\sin\theta\cos\theta=-\frac34 Divide by 2: sin⁡θcos⁡θ=−38\sin\theta\cos\theta=-\frac38 Now let's look at the traps. The choice 38\frac38 is what you get if you accidentally compute 1−14=341-\frac14=\frac34 instead of 14−1=−34\frac14-1=-\frac34, losing the negative sign. The choice −34-\frac34 comes from solving 2sin⁡θcos⁡θ=−342\sin\theta\cos\theta=-\frac34 but forgetting to divide by 2. The choice 14\frac14 treats sin⁡θcos⁡θ\sin\theta\cos\theta as if it were equal to (sin⁡θ+cos⁡θ)2(\sin\theta+\cos\theta)^2, which confuses the sum with the product. Your study tip: whenever a problem gives sin⁡θ+cos⁡θ\sin\theta+\cos\theta or sin⁡θ−cos⁡θ\sin\theta-\cos\theta and asks for a product like sin⁡θcos⁡θ\sin\theta\cos\theta, square the sum immediately. Then use sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 and carefully track the negative sign when moving 11 across the equation.

Question 8

Given sin⁡θ=35\sin \theta=\frac{3}{5} with θ\theta in Quadrant II, evaluate tan⁡θcot⁡θ+cos⁡2θ\tan \theta\cot \theta+\cos^2 \theta.

  1. 11
  2. −1625-\frac{16}{25}
  3. 1625\frac{16}{25}
  4. 4125\frac{41}{25} (correct answer)
Explanation: Whenever you see a trig expression involving identities, simplify the structure before plugging in numbers. Here the key simplification is tan⁡θcot⁡θ=1\tan\theta\cot\theta=1, since cot⁡θ=1tan⁡θ\cot\theta=\frac{1}{\tan\theta}. So the first term contributes exactly 11, regardless of the quadrant. The remaining part uses the Pythagorean identity: sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1. Given sin⁡θ=35\sin\theta=\frac{3}{5}, you get cos⁡2θ=1−925=1625\cos^2\theta=1-\frac{9}{25}=\frac{16}{25}. Quadrant II tells you cos⁡θ\cos\theta is negative, so cos⁡θ=−45\cos\theta=-\frac45, but cos⁡2θ\cos^2\theta is still positive 1625\frac{16}{25}. Therefore the whole expression is 1+1625=4125.1+\frac{16}{25}=\frac{41}{25}. The choice 1625\frac{16}{25} is only the cosine-squared piece; it forgets the tan⁡θcot⁡θ=1\tan\theta\cot\theta=1 term. The choice 11 is only the product identity; it ignores the cosine-squared term entirely. The choice −1625-\frac{16}{25} likely comes from carrying the negative sign of cos⁡θ=−45\cos\theta=-\frac45 into the square, but a squared value can never be negative. Study tip: first simplify identities like tan⁡θcot⁡θ\tan\theta\cot\theta, then use Pythagorean identities. Use the quadrant only to determine signs of unsquared trig functions, never to make a square negative.

Question 9

Which equation is NOT an identity for all values of xx for which both sides are defined?

  1. tan⁡xsec⁡x=1csc⁡x\frac{\tan x}{\sec x}=\frac{1}{\csc x}
  2. cot⁡xcsc⁡x=1sec⁡x\frac{\cot x}{\csc x}=\frac{1}{\sec x}
  3. cos⁡x1−sin⁡x=1−sin⁡xcos⁡x\frac{\cos x}{1-\sin x}=\frac{1-\sin x}{\cos x} (correct answer)
  4. cos⁡x1−sin⁡x=1+sin⁡xcos⁡x\frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x}
Explanation: Whenever you see a trigonometric identity question, rewrite expressions in terms of sine and cosine, or cross-multiply to test whether the two sides are truly equivalent. An identity must hold for every value in the domain, not just a few special angles. Check tan⁡xsec⁡x\frac{\tan x}{\sec x}: since tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x} and sec⁡x=1cos⁡x\sec x=\frac{1}{\cos x}, this becomes sin⁡x\sin x, which equals 1csc⁡x\frac{1}{\csc x}. True. Similarly, cot⁡xcsc⁡x=cos⁡x/sin⁡x1/sin⁡x=cos⁡x\frac{\cot x}{\csc x}=\frac{\cos x/\sin x}{1/\sin x}=\cos x, which equals 1sec⁡x\frac{1}{\sec x}. True. Now look at cos⁡x1−sin⁡x=1−sin⁡xcos⁡x\frac{\cos x}{1-\sin x}=\frac{1-\sin x}{\cos x}. Cross-multiplying gives cos⁡2x=(1−sin⁡x)2\cos^2 x=(1-\sin x)^2. Expanding the right side gives 1−2sin⁡x+sin⁡2x1-2\sin x+\sin^2 x, so the equation would require 2sin⁡x(sin⁡x−1)=02\sin x(\sin x-1)=0. That only works for special angles, not all values, so this is the equation that is NOT an identity. Compare with cos⁡x1−sin⁡x=1+sin⁡xcos⁡x\frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x}. Cross-multiplying gives cos⁡2x=(1−sin⁡x)(1+sin⁡x)=1−sin⁡2x=cos⁡2x\cos^2 x=(1-\sin x)(1+\sin x)=1-\sin^2 x=\cos^2 x, which is true wherever both sides are defined. This is the standard conjugate identity. Study tip: when you see 1−sin⁡x1-\sin x in a denominator, try multiplying by the conjugate 1+sin⁡x1+\sin x. And when verifying identities, cross-multiplying quickly reveals whether two rational trig expressions are genuinely equal.

Question 10

Given sec⁡θ−tan⁡θ=13\sec \theta-\tan \theta=\frac{1}{3} for an acute angle θ\theta, find sec⁡θ+tan⁡θ\sec \theta+\tan \theta.

  1. 13\frac{1}{3}
  2. 33 (correct answer)
  3. 19\frac{1}{9}
  4. 99
Explanation: Whenever you see a pairing like sec⁡θ−tan⁡θ\sec\theta-\tan\theta and sec⁡θ+tan⁡θ\sec\theta+\tan\theta, think about multiplying conjugates. The key identity is (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=sec⁡2θ−tan⁡2θ=1,(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=\sec^2\theta-\tan^2\theta=1, because sec⁡2θ=1+tan⁡2θ\sec^2\theta=1+\tan^2\theta. So the product of these two expressions is always exactly 11. Here you are told the first factor is 13\frac{1}{3}, so 13(sec⁡θ+tan⁡θ)=1⇒sec⁡θ+tan⁡θ=3.\frac{1}{3}(\sec\theta+\tan\theta)=1 \quad\Rightarrow\quad \sec\theta+\tan\theta=3. Since θ\theta is acute, both secant and tangent are positive, so the positive value 33 is correct. Why the wrong choices miss the mark: choosing 13\frac{1}{3} simply repeats the given value without using the conjugate relationship. Choosing 19\frac{1}{9} would be squaring the given value, but the identity calls for multiplication by the conjugate, not squaring. Choosing 99 comes from inverting the square of the given value, (13)−2=9\left(\frac{1}{3}\right)^{-2}=9, which incorrectly treats the product as 19\frac{1}{9} instead of 11. Study tip: whenever you see sec⁡θ±tan⁡θ\sec\theta\pm\tan\theta, immediately recall (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1.(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1. Treat one expression as an unknown factor and solve for it directly. This identity saves you from needing any trigonometric values and is a favorite shortcut on algebra exams.

Question 11

Given sec⁡θ−tan⁡θ=13\sec \theta-\tan \theta=\frac{1}{3} for an acute angle θ\theta, find sec⁡θ+tan⁡θ\sec \theta+\tan \theta.

  1. 13\frac{1}{3}
  2. 33 (correct answer)
  3. 19\frac{1}{9}
  4. 99
Explanation: Whenever you see sec⁡θ\sec\theta and tan⁡θ\tan\theta in the same equation, your first thought should be the Pythagorean identity in factored form: sec⁡2θ−tan⁡2θ=1\sec^2\theta-\tan^2\theta=1. Since sec⁡2θ−tan⁡2θ=(sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)\sec^2\theta-\tan^2\theta=(\sec\theta-\tan\theta)(\sec\theta+\tan\theta), the product of the two conjugate expressions is always 11. You're given that sec⁡θ−tan⁡θ=13\sec\theta-\tan\theta=\frac{1}{3}. Therefore, (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1 so sec⁡θ+tan⁡θ=11/3=3.\sec\theta+\tan\theta=\frac{1}{1/3}=3. The choice 13\frac{1}{3} is just the original difference, not the sum;treating them as equal would be a trap. The choice 19\frac{1}{9}is the square of the given difference, which has no place here unless you mistakenly think the product equals the square of one factor. The choice 99is the reciprocal of 19\frac{1}{9},and would only result from an incorrect squared version of the identity. . Since θ\thetais acute, sec⁡θ\sec\thetaandtan⁡θ\tan\thetaare positive, so the sum is definitely positive;indeed, the identity gives a unique value without needing to solve for θ\theta separately. . When you see a sec/tan pair, remember the conjugate-product identity: (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1.(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1. If you know one factor, the other is simply its reciprocal.

Question 12

Simplify (sec⁡x−cos⁡x)cot⁡x(\sec x-\cos x)\cot x for all values of xx for which the expression is defined.

  1. cos⁡x\cos x
  2. sec⁡x\sec x
  3. sin⁡xtan⁡x\sin x\tan x
  4. sin⁡x\sin x (correct answer)
Explanation: When you see a product of trig expressions, a good first move is to rewrite everything in terms of sine and cosine, or to look for a Pythagorean identity. Here, sec⁡x−cos⁡x=1cos⁡x−cos⁡x=1−cos⁡2xcos⁡x=sin⁡2xcos⁡x=sin⁡xtan⁡x.\sec x-\cos x=\frac{1}{\cos x}-\cos x=\frac{1-\cos^2 x}{\cos x}=\frac{\sin^2 x}{\cos x}=\sin x\tan x. Then multiply by cot⁡x=cos⁡xsin⁡x\cot x=\frac{\cos x}{\sin x}: (sec⁡x−cos⁡x)cot⁡x=(sin⁡xtan⁡x)cot⁡x=sin⁡x⋅(tan⁡xcot⁡x)=sin⁡x,(\sec x-\cos x)\cot x=(\sin x\tan x)\cot x=\sin x\cdot(\tan x\cot x)=\sin x, since tan⁡xcot⁡x=1\tan x\cot x=1. For all xx where the original expression is defined, this simplification is valid. The choice sin⁡xtan⁡x\sin x\tan x is a tempting intermediate, but it is what sec⁡x−cos⁡x\sec x-\cos x alone simplifies to—you still have the cot⁡x\cot x factor to use. The choice cos⁡x\cos x often comes from incorrectly reducing sec⁡x−cos⁡x\sec x-\cos x to just sin⁡x\sin x, then using sin⁡xcot⁡x=cos⁡x\sin x\cot x=\cos x; however, the missing tan⁡x\tan x factor is exactly what cancels with cot⁡x\cot x. The choice sec⁡x\sec x would mean dropping the cot⁡x\cot x factor entirely or treating it as 11, but cot⁡x\cot x must be multiplied through. As a study tip: rewrite trig expressions in sine and cosine, simplify fully, and check whether an answer choice is only an intermediate step. On this exam, many wrong answers are "stopping too early" traps.

Question 13

Which expression is equivalent to sin⁡4x−cos⁡4xsin⁡2x−cos⁡2x\frac{\sin^4 x-\cos^4 x}{\sin^2 x-\cos^2 x} on its domain?

  1. −(sin⁡2x+cos⁡2x)-(\sin^2 x+\cos^2 x)
  2. 00
  3. sin⁡2x+cos⁡2x\sin^2 x+\cos^2 x (correct answer)
  4. sin⁡2x−cos⁡2x\sin^2 x-\cos^2 x
Explanation: Factor the numerator as a difference of squares: sin⁡4x−cos⁡4x=(sin⁡2x−cos⁡2x)(sin⁡2x+cos⁡2x)\sin^4 x-\cos^4 x=(\sin^2 x-\cos^2 x)(\sin^2 x+\cos^2 x). Canceling sin⁡2x−cos⁡2x\sin^2 x-\cos^2 x leaves sin⁡2x+cos⁡2x\sin^2 x+\cos^2 x. The negative form is a sign error, zero comes from canceling incorrectly, and sin⁡2x−cos⁡2x\sin^2 x-\cos^2 x mistakes the denominator for the simplified quotient.

Question 14

If sec⁡θ+tan⁡θ=4\sec\theta+\tan\theta=4, what is sec⁡θ−tan⁡θ\sec\theta-\tan\theta?

  1. 14\frac14 (correct answer)
  2. 12\frac12
  3. 22
  4. 44
Explanation: Multiply the two expressions: (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=sec⁡2θ−tan⁡2θ=1(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=\sec^2\theta-\tan^2\theta=1. Since the first factor is 4, the second factor must be 14\frac14. The other choices come from taking a square root, halving, or assuming the difference equals the given sum.

Question 15

Which expression is NOT equivalent to sec⁡xcsc⁡x\sec x\csc x on its domain?

  1. tan⁡x+cot⁡x\tan x+\cot x
  2. sec⁡2xtan⁡x\frac{\sec^2 x}{\tan x}
  3. csc⁡2xcot⁡x\frac{\csc^2 x}{\cot x}
  4. cot⁡xsin⁡2x\frac{\cot x}{\sin^2 x} (correct answer)
Explanation: The sum tan⁡x+cot⁡x=sin⁡xcos⁡x+cos⁡xsin⁡x=1sin⁡xcos⁡x=sec⁡xcsc⁡x\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{1}{\sin x\cos x}=\sec x\csc x. Also, sec⁡2xtan⁡x=1cos⁡2x⋅cos⁡xsin⁡x=sec⁡xcsc⁡x\frac{\sec^2 x}{\tan x}=\frac{1}{\cos^2 x}\cdot\frac{\cos x}{\sin x}=\sec x\csc x, and csc⁡2xcot⁡x=1sin⁡2x⋅sin⁡xcos⁡x=sec⁡xcsc⁡x\frac{\csc^2 x}{\cot x}=\frac{1}{\sin^2 x}\cdot\frac{\sin x}{\cos x}=\sec x\csc x. The expression with cotangent in the numerator, cot⁡xsin⁡2x=cos⁡xsin⁡3x\frac{\cot x}{\sin^2 x}=\frac{\cos x}{\sin^3 x}, is not sec⁡xcsc⁡x\sec x\csc x.

Question 16

Which expression is equivalent to cos⁡2xcsc⁡x−1\frac{\cos^2 x}{\csc x-1} on its domain?

  1. sin⁡x(1+sin⁡x)\sin x(1+\sin x) (correct answer)
  2. sin⁡x(1−sin⁡x)\sin x(1-\sin x)
  3. −sin⁡x(1+sin⁡x)-\sin x(1+\sin x)
  4. sin⁡xcsc⁡x+1\frac{\sin x}{\csc x+1}
Explanation: Since csc⁡x−1=1sin⁡x−1=1−sin⁡xsin⁡x\csc x-1=\frac{1}{\sin x}-1=\frac{1-\sin x}{\sin x}, the given expression is cos⁡2x⋅sin⁡x1−sin⁡x\cos^2 x\cdot\frac{\sin x}{1-\sin x}. Using cos⁡2x=1−sin⁡2x=(1−sin⁡x)(1+sin⁡x)\cos^2 x=1-\sin^2 x=(1-\sin x)(1+\sin x) gives sin⁡x(1+sin⁡x)\sin x(1+\sin x). The expression sin⁡x(1−sin⁡x)\sin x(1-\sin x) uses the wrong factor after cancellation; −sin⁡x(1+sin⁡x)-\sin x(1+\sin x) has a sign error; and sin⁡xcsc⁡x+1\frac{\sin x}{\csc x+1} inverts the denominator incorrectly.

Question 17

If sec⁡2x−1csc⁡2x−1=ktan⁡4x\frac{\sec^2 x-1}{\csc^2 x-1}=k\tan^4 x for all xx in the domain, what is kk?

  1. −1-1
  2. 11 (correct answer)
  3. 22
  4. 44
Explanation: Use sec⁡2x−1=tan⁡2x\sec^2 x-1=\tan^2 x and csc⁡2x−1=cot⁡2x\csc^2 x-1=\cot^2 x. The quotient is tan⁡2xcot⁡2x=tan⁡2x⋅tan⁡2x=tan⁡4x\frac{\tan^2 x}{\cot^2 x}=\tan^2 x\cdot\tan^2 x=\tan^4 x, so k=1k=1. The other choices come from sign errors or from treating the quotient as 2tan⁡4x2\tan^4 x or 4tan⁡4x4\tan^4 x.

Question 18

If tan⁡θ+cot⁡θ=3\tan\theta+\cot\theta=3, what is tan⁡2θ+cot⁡2θ\tan^2\theta+\cot^2\theta?

  1. 55
  2. 77 (correct answer)
  3. 99
  4. 1111
Explanation: Square the given equation: (tan⁡θ+cot⁡θ)2=9(\tan\theta+\cot\theta)^2=9. Expanding gives tan⁡2θ+cot⁡2θ+2tan⁡θcot⁡θ=9\tan^2\theta+\cot^2\theta+2\tan\theta\cot\theta=9. Since tan⁡θcot⁡θ=1\tan\theta\cot\theta=1, tan⁡2θ+cot⁡2θ+2=9\tan^2\theta+\cot^2\theta+2=9, so tan⁡2θ+cot⁡2θ=7\tan^2\theta+\cot^2\theta=7. The value 99 omits the product term; the values 55 and 1111 use the product term with an incorrect sign or coefficient.

Question 19

If sin⁡θ+cos⁡θ=12\sin\theta+\cos\theta=\frac12, what is sin⁡θcos⁡θ\sin\theta\cos\theta?

  1. −34-\frac34
  2. −38-\frac38 (correct answer)
  3. 38\frac38
  4. 34\frac34
Explanation: Square both sides: (sin⁡θ+cos⁡θ)2=14(\sin\theta+\cos\theta)^2=\frac14. Expanding gives sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=14\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=\frac14, so 1+2sin⁡θcos⁡θ=141+2\sin\theta\cos\theta=\frac14. Thus 2sin⁡θcos⁡θ=−342\sin\theta\cos\theta=-\frac34, and sin⁡θcos⁡θ=−38\sin\theta\cos\theta=-\frac38. The value −34-\frac34 comes from forgetting to divide by 2; the positive values have the wrong sign.

Question 20

If sec⁡θ=−3\sec\theta=-3 and θ\theta lies in Quadrant II, what is tan⁡θ\tan\theta?

  1. −22-2\sqrt{2} (correct answer)
  2. −223-\frac{2\sqrt{2}}{3}
  3. 223\frac{2\sqrt{2}}{3}
  4. 222\sqrt{2}
Explanation: Since sec⁡θ=−3\sec\theta=-3, cos⁡θ=−13\cos\theta=-\frac13. Use 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta, so tan⁡2θ=9−1=8\tan^2\theta=9-1=8. In Quadrant II, tangent is negative, so tan⁡θ=−8=−22\tan\theta=-\sqrt{8}=-2\sqrt{2}. The positive radical has the wrong sign, and the denominator-3 choices confuse tangent with a sine/cosine ratio.