Math 3 Quiz: Trigonometric Identities
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Trigonometric IdentitiesQuestion 1 of 15

If cot⁡γ=−724\cot \gamma = -\frac{7}{24} and sin⁡γ>0\sin \gamma > 0, what is the value of csc⁡γ\csc \gamma?

−257-\frac{25}{7}
257\frac{25}{7}
−2524-\frac{25}{24}
2524\frac{25}{24}
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Math 3 Quiz

Math 3 Quiz: Trigonometric Identities

Practice Trigonometric Identities in Math 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 Trigonometric Identities, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 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

If cot⁡γ=−724\cot \gamma = -\frac{7}{24} and sin⁡γ>0\sin \gamma > 0, what is the value of csc⁡γ\csc \gamma?

  1. −257-\frac{25}{7}
  2. 257\frac{25}{7}
  3. −2524-\frac{25}{24}
  4. 2524\frac{25}{24} (correct answer)
Explanation: When you encounter trigonometric problems with given ratios and quadrant constraints, you need to use the Pythagorean identity and determine signs based on which quadrant the angle lies in. Since cot⁡γ=−724\cot \gamma = -\frac{7}{24} and sin⁡γ>0\sin \gamma > 0, you can determine that γ\gamma is in Quadrant II. Here's why: cotangent equals cos⁡γsin⁡γ\frac{\cos \gamma}{\sin \gamma}, so if cotangent is negative and sine is positive, then cosine must be negative—which only occurs in Quadrant II. To find csc⁡γ=1sin⁡γ\csc \gamma = \frac{1}{\sin \gamma}, you first need to find sin⁡γ\sin \gamma. Using the identity cot⁡2γ+1=csc⁡2γ\cot^2 \gamma + 1 = \csc^2 \gamma: (−724)2+1=csc⁡2γ\left(-\frac{7}{24}\right)^2 + 1 = \csc^2 \gamma 49576+1=49+576576=625576=csc⁡2γ\frac{49}{576} + 1 = \frac{49 + 576}{576} = \frac{625}{576} = \csc^2 \gamma Therefore, csc⁡γ=±2524\csc \gamma = \pm\frac{25}{24}. Since sin⁡γ>0\sin \gamma > 0 in Quadrant II, csc⁡γ=1sin⁡γ>0\csc \gamma = \frac{1}{\sin \gamma} > 0, so csc⁡γ=2524\csc \gamma = \frac{25}{24}. Choice A (−257-\frac{25}{7}) and choice B (257\frac{25}{7}) incorrectly use 7 in the denominator, likely from confusing cotangent's numerator with the final answer's denominator. Choice C (−2524-\frac{25}{24}) has the correct magnitude but wrong sign—this would be correct if the angle were in Quadrant III where both sine and cosecant are negative. Remember: always check quadrant signs after calculating trigonometric ratios. The given conditions about positive or negative values tell you which quadrant you're in, which determines the final sign.

Question 2

Which expression is equivalent to cos⁡2x−sin⁡2xcos⁡2x+sin⁡2x\frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x} for all values of xx where the expression is defined?

  1. cos⁡2x\cos 2x (correct answer)
  2. sin⁡2x\sin 2x
  3. tan⁡2x\tan^2 x
  4. sec⁡2x\sec^2 x
Explanation: Using the fundamental identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, the denominator becomes cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1. The numerator cos⁡2x−sin⁡2x\cos^2 x - \sin^2 x is the double angle identity for cosine: cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x. Therefore, the expression simplifies to cos⁡2x1=cos⁡2x\frac{\cos 2x}{1} = \cos 2x. Choice B is wrong because sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x. Choice C is wrong because tan⁡2x=sin⁡2xcos⁡2x\tan^2 x = \frac{\sin^2 x}{\cos^2 x}. Choice D is wrong because sec⁡2x=1cos⁡2x\sec^2 x = \frac{1}{\cos^2 x}.

Question 3

If cos⁡α=513\cos \alpha = \frac{5}{13} and sin⁡α>0\sin \alpha > 0, what is the value of 1+tan⁡2αsec⁡α\frac{1 + \tan^2 \alpha}{\sec \alpha}?

  1. 135\frac{13}{5} (correct answer)
  2. 513\frac{5}{13}
  3. 16925\frac{169}{25}
  4. 25169\frac{25}{169}
Explanation: First, we use the identity 1+tan⁡2α=sec⁡2α1 + \tan^2 \alpha = \sec^2 \alpha. So the expression becomes sec⁡2αsec⁡α=sec⁡α\frac{\sec^2 \alpha}{\sec \alpha} = \sec \alpha. Since cos⁡α=513\cos \alpha = \frac{5}{13}, we have sec⁡α=1cos⁡α=135\sec \alpha = \frac{1}{\cos \alpha} = \frac{13}{5}. Choice B gives cos⁡α\cos \alpha instead of sec⁡α\sec \alpha. Choice C would result from squaring sec⁡α\sec \alpha. Choice D would result from squaring cos⁡α\cos \alpha.

Question 4

If cos⁡α=513\cos \alpha = \frac{5}{13} and α\alpha is in the fourth quadrant, what is the value of sin⁡(2α)\sin(2\alpha)?

  1. −120169-\frac{120}{169} (correct answer)
  2. −119169-\frac{119}{169}
  3. 119169\frac{119}{169}
  4. 120169\frac{120}{169}
Explanation: First, we find sin⁡α\sin \alpha using sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1. We have sin⁡2α=1−25169=144169\sin^2 \alpha = 1 - \frac{25}{169} = \frac{144}{169}, so sin⁡α=±1213\sin \alpha = \pm\frac{12}{13}. Since α\alpha is in quadrant IV, sin⁡α=−1213\sin \alpha = -\frac{12}{13}. Using the double angle identity sin⁡(2α)=2sin⁡αcos⁡α\sin(2\alpha) = 2\sin \alpha \cos \alpha, we get sin⁡(2α)=2⋅(−1213)⋅513=−120169\sin(2\alpha) = 2 \cdot (-\frac{12}{13}) \cdot \frac{5}{13} = -\frac{120}{169}. Choice B makes an arithmetic error in the calculation. Choice C uses the wrong sign for sin⁡α\sin \alpha. Choice D makes both sign and calculation errors.

Question 5

If tan⁡θ=−43\tan \theta = -\frac{4}{3} and sin⁡θ>0\sin \theta > 0, what is the value of sin⁡θ−cos⁡θ\sin \theta - \cos \theta?

  1. 75\frac{7}{5} (correct answer)
  2. 15\frac{1}{5}
  3. −15-\frac{1}{5}
  4. −75-\frac{7}{5}
Explanation: Since tan⁡θ=−43<0\tan \theta = -\frac{4}{3} < 0 and sin⁡θ>0\sin \theta > 0, we must have cos⁡θ<0\cos \theta < 0, so θ\theta is in quadrant II. Using tan⁡θ=sin⁡θcos⁡θ=−43\tan \theta = \frac{\sin \theta}{\cos \theta} = -\frac{4}{3} and sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, we can write sin⁡θ=−43cos⁡θ\sin \theta = -\frac{4}{3}\cos \theta. Substituting: (−43cos⁡θ)2+cos⁡2θ=1(-\frac{4}{3}\cos \theta)^2 + \cos^2 \theta = 1, so 169cos⁡2θ+cos⁡2θ=1\frac{16}{9}\cos^2 \theta + \cos^2 \theta = 1, giving 259cos⁡2θ=1\frac{25}{9}\cos^2 \theta = 1. Thus cos⁡2θ=925\cos^2 \theta = \frac{9}{25}, so cos⁡θ=±35\cos \theta = \pm\frac{3}{5}. Since we're in quadrant II, cos⁡θ=−35\cos \theta = -\frac{3}{5} and sin⁡θ=−43⋅(−35)=45\sin \theta = -\frac{4}{3} \cdot (-\frac{3}{5}) = \frac{4}{5}. Therefore, sin⁡θ−cos⁡θ=45−(−35)=45+35=75\sin \theta - \cos \theta = \frac{4}{5} - (-\frac{3}{5}) = \frac{4}{5} + \frac{3}{5} = \frac{7}{5}. Choice B subtracts incorrectly. Choice C uses wrong signs. Choice D uses both terms negative.

Question 6

For which expression is sin⁡4ξ+cos⁡4ξ\sin^4 \xi + \cos^4 \xi equivalent?

  1. 1−sin⁡2ξcos⁡2ξ1 - \sin^2 \xi \cos^2 \xi
  2. 1−2sin⁡2ξcos⁡2ξ1 - 2\sin^2 \xi \cos^2 \xi (correct answer)
  3. 1+sin⁡2ξcos⁡2ξ1 + \sin^2 \xi \cos^2 \xi
  4. 1+2sin⁡2ξcos⁡2ξ1 + 2\sin^2 \xi \cos^2 \xi
Explanation: We can use the identity (sin⁡2ξ+cos⁡2ξ)2=1(\sin^2 \xi + \cos^2 \xi)^2 = 1 and expand: sin⁡4ξ+2sin⁡2ξcos⁡2ξ+cos⁡4ξ=1\sin^4 \xi + 2\sin^2 \xi \cos^2 \xi + \cos^4 \xi = 1. Rearranging: sin⁡4ξ+cos⁡4ξ=1−2sin⁡2ξcos⁡2ξ\sin^4 \xi + \cos^4 \xi = 1 - 2\sin^2 \xi \cos^2 \xi. Choice A has the wrong coefficient. Choice C has the wrong sign. Choice D has both wrong sign and coefficient.

Question 7

For which value of kk does the equation sin⁡2θ+kcos⁡2θ=1\sin^2 \theta + k\cos^2 \theta = 1 hold for all values of θ\theta?

  1. k=0k = 0
  2. k=1k = 1 (correct answer)
  3. k=−1k = -1
  4. k=2k = 2
Explanation: Using the fundamental identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, we can substitute to get sin⁡2θ+kcos⁡2θ=1\sin^2 \theta + k\cos^2 \theta = 1. For this to equal the fundamental identity, we need kcos⁡2θ=cos⁡2θk\cos^2 \theta = \cos^2 \theta, which means k=1k = 1. Choice A would give sin⁡2θ=1\sin^2 \theta = 1, which is not true for all θ\theta. Choice C would give sin⁡2θ−cos⁡2θ=1\sin^2 \theta - \cos^2 \theta = 1, which is not always true. Choice D would give sin⁡2θ+2cos⁡2θ=1\sin^2 \theta + 2\cos^2 \theta = 1, which contradicts the fundamental identity.

Question 8

For what value of cos⁡ϕ\cos \phi does the expression sin⁡2ϕ+cos⁡2ϕcos⁡ϕ\frac{\sin^2 \phi + \cos^2 \phi}{\cos \phi} equal 43\frac{4}{3}?

  1. cos⁡ϕ=13\cos \phi = \frac{1}{3}
  2. cos⁡ϕ=43\cos \phi = \frac{4}{3}
  3. cos⁡ϕ=34\cos \phi = \frac{3}{4} (correct answer)
  4. cos⁡ϕ=14\cos \phi = \frac{1}{4}
Explanation: When you encounter trigonometric expressions with both sine and cosine terms, always look for opportunities to apply fundamental identities. The key insight here is recognizing that sin⁡2ϕ+cos⁡2ϕ=1\sin^2 \phi + \cos^2 \phi = 1 by the Pythagorean identity. Let's simplify the given expression first. Since sin⁡2ϕ+cos⁡2ϕ=1\sin^2 \phi + \cos^2 \phi = 1, we can rewrite: sin⁡2ϕ+cos⁡2ϕcos⁡ϕ=1cos⁡ϕ\frac{\sin^2 \phi + \cos^2 \phi}{\cos \phi} = \frac{1}{\cos \phi} Now we need to solve: 1cos⁡ϕ=43\frac{1}{\cos \phi} = \frac{4}{3} Cross-multiplying gives us: 3=4cos⁡ϕ3 = 4\cos \phi Therefore: cos⁡ϕ=34\cos \phi = \frac{3}{4} This confirms that answer C is correct. Let's examine why the other options fail. If we substitute each into our equation 1cos⁡ϕ=43\frac{1}{\cos \phi} = \frac{4}{3}: For A) cos⁡ϕ=13\cos \phi = \frac{1}{3}: This gives 113=3≠43\frac{1}{\frac{1}{3}} = 3 \neq \frac{4}{3} For B) cos⁡ϕ=43\cos \phi = \frac{4}{3}: This gives 143=34≠43\frac{1}{\frac{4}{3}} = \frac{3}{4} \neq \frac{4}{3}. Also, cosine values must be between -1 and 1, so 43\frac{4}{3} is impossible. For D) cos⁡ϕ=14\cos \phi = \frac{1}{4}: This gives 114=4≠43\frac{1}{\frac{1}{4}} = 4 \neq \frac{4}{3} Study tip: Always simplify trigonometric expressions using basic identities before solving equations. The Pythagorean identity sin⁡2ϕ+cos⁡2ϕ=1\sin^2 \phi + \cos^2 \phi = 1 is especially powerful for eliminating complex terms and revealing simpler relationships.

Question 9

If sec⁡2δ−tan⁡2δ=k\sec^2 \delta - \tan^2 \delta = k for all values of δ\delta where the expression is defined, what is the value of kk?

  1. k=0k = 0
  2. k=1k = 1 (correct answer)
  3. k=−1k = -1
  4. k=2k = 2
Explanation: This is a fundamental trigonometric identity: sec⁡2δ−tan⁡2δ=1\sec^2 \delta - \tan^2 \delta = 1. This can be derived from sin⁡2δ+cos⁡2δ=1\sin^2 \delta + \cos^2 \delta = 1 by dividing both sides by cos⁡2δ\cos^2 \delta: sin⁡2δcos⁡2δ+cos⁡2δcos⁡2δ=1cos⁡2δ\frac{\sin^2 \delta}{\cos^2 \delta} + \frac{\cos^2 \delta}{\cos^2 \delta} = \frac{1}{\cos^2 \delta}, which gives tan⁡2δ+1=sec⁡2δ\tan^2 \delta + 1 = \sec^2 \delta, or sec⁡2δ−tan⁡2δ=1\sec^2 \delta - \tan^2 \delta = 1. Choice A would contradict the fundamental identity. Choice C represents tan⁡2δ−sec⁡2δ\tan^2 \delta - \sec^2 \delta. Choice D would be sec⁡2δ+tan⁡2δ\sec^2 \delta + \tan^2 \delta minus something.

Question 10

If cos⁡ψ=−35\cos \psi = -\frac{3}{5} and ψ\psi is in the second quadrant, what is the value of sin⁡ψ1+cos⁡ψ\frac{\sin \psi}{1 + \cos \psi}?

  1. 12\frac{1}{2}
  2. −2-2
  3. 22 (correct answer)
  4. −12-\frac{1}{2}
Explanation: When you encounter trigonometric problems with given values and quadrant information, your first step is to find the missing trigonometric function using the Pythagorean identity, then carefully track signs based on the quadrant. Since cos⁡ψ=−35\cos \psi = -\frac{3}{5} and ψ\psi is in the second quadrant, you need to find sin⁡ψ\sin \psi. Using sin⁡2ψ+cos⁡2ψ=1\sin^2 \psi + \cos^2 \psi = 1: sin⁡2ψ=1−cos⁡2ψ=1−(−35)2=1−925=1625\sin^2 \psi = 1 - \cos^2 \psi = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} So sin⁡ψ=±45\sin \psi = \pm\frac{4}{5}. Since ψ\psi is in the second quadrant where sine is positive, sin⁡ψ=45\sin \psi = \frac{4}{5}. Now you can evaluate the expression: sin⁡ψ1+cos⁡ψ=451+(−35)=4525=45⋅52=2\frac{\sin \psi}{1 + \cos \psi} = \frac{\frac{4}{5}}{1 + \left(-\frac{3}{5}\right)} = \frac{\frac{4}{5}}{\frac{2}{5}} = \frac{4}{5} \cdot \frac{5}{2} = 2 The correct answer is C) 22. Choice A) 12\frac{1}{2} results from incorrectly using sin⁡ψ=15\sin \psi = \frac{1}{5} instead of the correct value. Choice B) −2-2 comes from using the wrong sign for sine (treating it as negative in the second quadrant). Choice D) −12-\frac{1}{2} combines both errors: wrong sine value and wrong sign. Remember: in quadrant problems, always determine signs carefully using "All Students Take Calculus" (All positive in I, Sine positive in II, Tangent positive in III, Cosine positive in IV).

Question 11

Which of the following is equivalent to sin⁡2β1−cos⁡β\frac{\sin^2 \beta}{1 - \cos \beta} for cos⁡β≠1\cos \beta \neq 1?

  1. 1+cos⁡β1 + \cos \beta (correct answer)
  2. 1−cos⁡β1 - \cos \beta
  3. 11+cos⁡β\frac{1}{1 + \cos \beta}
  4. 11−cos⁡β\frac{1}{1 - \cos \beta}
Explanation: Using sin⁡2β=1−cos⁡2β=(1−cos⁡β)(1+cos⁡β)\sin^2 \beta = 1 - \cos^2 \beta = (1 - \cos \beta)(1 + \cos \beta), we can substitute: sin⁡2β1−cos⁡β=(1−cos⁡β)(1+cos⁡β)1−cos⁡β=1+cos⁡β\frac{\sin^2 \beta}{1 - \cos \beta} = \frac{(1 - \cos \beta)(1 + \cos \beta)}{1 - \cos \beta} = 1 + \cos \beta (provided cos⁡β≠1\cos \beta \neq 1). Choice B is the denominator. Choice C would result from incorrectly inverting the entire expression. Choice D would result from incorrectly canceling terms.

Question 12

Given that cos⁡λ≠0\cos \lambda \neq 0 and sin⁡λ≠0\sin \lambda \neq 0, which of the following trigonometric identities must be true?

  1. tan⁡2λ+1=sec⁡2λ\tan^2 \lambda + 1 = \sec^2 \lambda
  2. cot⁡2λ+1=csc⁡2λ\cot^2 \lambda + 1 = \csc^2 \lambda
  3. sec⁡2λ−tan⁡2λ=1\sec^2 \lambda - \tan^2 \lambda = 1
  4. All of the above (correct answer)
Explanation: These are fundamental trigonometric identities derived from sin⁡2λ+cos⁡2λ=1\sin^2 \lambda + \cos^2 \lambda = 1. Identity A: Dividing by cos⁡2λ\cos^2 \lambda gives tan⁡2λ+1=sec⁡2λ\tan^2 \lambda + 1 = \sec^2 \lambda. Identity B: Dividing by sin⁡2λ\sin^2 \lambda gives 1+cot⁡2λ=csc⁡2λ1 + \cot^2 \lambda = \csc^2 \lambda. Identity C: This is equivalent to identity A. Since cos⁡λ≠0\cos \lambda \neq 0 and sin⁡λ≠0\sin \lambda \neq 0, all three identities are valid.

Question 13

Given that sin⁡A=513\sin A = \frac{5}{13} and cos⁡A=1213\cos A = \frac{12}{13}, what is the value of sin⁡2A+cos⁡2A+tan⁡2A\sin^2 A + \cos^2 A + \tan^2 A?

  1. 394169\frac{394}{169}
  2. 313169\frac{313}{169}
  3. 169144\frac{169}{144} (correct answer)
  4. 25144\frac{25}{144}
Explanation: When you encounter trigonometric expressions involving multiple functions, start by identifying what fundamental relationships you can use. This question tests your knowledge of the Pythagorean identity and how to find tangent from sine and cosine. To find sin⁡2A+cos⁡2A+tan⁡2A\sin^2 A + \cos^2 A + \tan^2 A, begin with what you know. The Pythagorean identity tells us that sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1 for any angle. This is always true regardless of the specific values. Next, find tan⁡2A\tan^2 A. Since tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}, we have: tan⁡A=5/1312/13=512\tan A = \frac{5/13}{12/13} = \frac{5}{12} Therefore: tan⁡2A=(512)2=25144\tan^2 A = \left(\frac{5}{12}\right)^2 = \frac{25}{144} The complete expression equals: sin⁡2A+cos⁡2A+tan⁡2A=1+25144=144144+25144=169144\sin^2 A + \cos^2 A + \tan^2 A = 1 + \frac{25}{144} = \frac{144}{144} + \frac{25}{144} = \frac{169}{144} Answer A (394169\frac{394}{169}) likely results from incorrectly calculating sin⁡2A+cos⁡2A\sin^2 A + \cos^2 A as (513)2+(1213)2=169169\left(\frac{5}{13}\right)^2 + \left(\frac{12}{13}\right)^2 = \frac{169}{169} and making arithmetic errors. Answer B (313169\frac{313}{169}) comes from similar calculation mistakes with the denominators. Answer D (25144\frac{25}{144}) represents finding only tan⁡2A\tan^2 A and forgetting to add the Pythagorean identity term. Remember: sin⁡2A+cos⁡2A\sin^2 A + \cos^2 A always equals 1, so focus your calculations on the additional trigonometric terms in these types of problems.

Question 14

If sec⁡α+tan⁡α=3\sec \alpha + \tan \alpha = 3, what is the value of sec⁡α−tan⁡α\sec \alpha - \tan \alpha?

  1. 13\frac{1}{3} (correct answer)
  2. 19\frac{1}{9}
  3. 33
  4. −13-\frac{1}{3}
Explanation: We use the identity (sec⁡α+tan⁡α)(sec⁡α−tan⁡α)=sec⁡2α−tan⁡2α=1(\sec \alpha + \tan \alpha)(\sec \alpha - \tan \alpha) = \sec^2 \alpha - \tan^2 \alpha = 1 (from the Pythagorean identity sec⁡2α−tan⁡2α=1\sec^2 \alpha - \tan^2 \alpha = 1). Given that sec⁡α+tan⁡α=3\sec \alpha + \tan \alpha = 3, we have 3⋅(sec⁡α−tan⁡α)=13 \cdot (\sec \alpha - \tan \alpha) = 1. Therefore, sec⁡α−tan⁡α=13\sec \alpha - \tan \alpha = \frac{1}{3}. Choice B results from incorrectly squaring the given equation. Choice C assumes the two expressions are equal. Choice D uses the wrong sign from a calculation error.

Question 15

Which of the following expressions is NOT equivalent to sin⁡x1+cos⁡x\frac{\sin x}{1 + \cos x}?

  1. 1−cos⁡xsin⁡x\frac{1 - \cos x}{\sin x}
  2. cot⁡(x2)−csc⁡(x2)\cot\left(\frac{x}{2}\right) - \csc\left(\frac{x}{2}\right) (correct answer)
  3. tan⁡(x2)\tan\left(\frac{x}{2}\right)
  4. 2sin⁡(x2)cos⁡(x2)2cos⁡2(x2)\frac{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{2\cos^2\left(\frac{x}{2}\right)}
Explanation: We can verify each choice. For choice A: multiply numerator and denominator of sin⁡x1+cos⁡x\frac{\sin x}{1 + \cos x} by (1−cos⁡x)(1 - \cos x) to get sin⁡x(1−cos⁡x)(1+cos⁡x)(1−cos⁡x)=sin⁡x(1−cos⁡x)1−cos⁡2x=sin⁡x(1−cos⁡x)sin⁡2x=1−cos⁡xsin⁡x\frac{\sin x(1 - \cos x)}{(1 + \cos x)(1 - \cos x)} = \frac{\sin x(1 - \cos x)}{1 - \cos^2 x} = \frac{\sin x(1 - \cos x)}{\sin^2 x} = \frac{1 - \cos x}{\sin x}. So A is equivalent. For choice C: using half-angle identities, sin⁡x1+cos⁡x=2sin⁡(x/2)cos⁡(x/2)1+(2cos⁡2(x/2)−1)=2sin⁡(x/2)cos⁡(x/2)2cos⁡2(x/2)=sin⁡(x/2)cos⁡(x/2)=tan⁡(x/2)\frac{\sin x}{1 + \cos x} = \frac{2\sin(x/2)\cos(x/2)}{1 + (2\cos^2(x/2) - 1)} = \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{\sin(x/2)}{\cos(x/2)} = \tan(x/2). So C is equivalent. Choice D simplifies to 2sin⁡(x/2)cos⁡(x/2)2cos⁡2(x/2)=sin⁡(x/2)cos⁡(x/2)=tan⁡(x/2)\frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} = \frac{\sin(x/2)}{\cos(x/2)} = \tan(x/2), which we showed equals the original expression. For choice B: cot⁡(x/2)−csc⁡(x/2)=cos⁡(x/2)sin⁡(x/2)−1sin⁡(x/2)=cos⁡(x/2)−1sin⁡(x/2)\cot(x/2) - \csc(x/2) = \frac{\cos(x/2)}{\sin(x/2)} - \frac{1}{\sin(x/2)} = \frac{\cos(x/2) - 1}{\sin(x/2)}. This is the negative of what we want, so B is NOT equivalent.