Fundamental Trig Identities - Algebra 3

1. Which expression is equivalent to $\frac{\cos^2 x}{\csc x-1}$ on its domain?
   Answer: $\sin x(1+\sin x)$ Since $\csc x-1=\frac{1}{\sin x}-1=\frac{1-\sin x}{\sin x}$, the given expression is $\cos^2 x\cdot\frac{\sin x}{1-\sin x}$. Using $\cos^2 x=1-\sin^2 x=(1-\sin x)(1+\sin x)$ gives $\sin x(1+\sin x)$. The expression $\sin x(1-\sin x)$ uses the wrong factor after cancellation; $-\sin x(1+\sin x)$ has a sign error; and $\frac{\sin x}{\csc x+1}$ inverts the denominator incorrectly.
2. Simplify $(\sec x-\cos x)\cot x$ for all values of $x$ for which the expression is defined.
   Answer: $\sin x$ 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, $$\sec x-\cos x=\frac{1}{\cos x}-\cos x=\frac{1-\cos^2 x}{\cos x}=\frac{\sin^2 x}{\cos x}=\sin x\tan x.$$ Then multiply by $\cot x=\frac{\cos x}{\sin x}$: $$(\sec x-\cos x)\cot x=(\sin x\tan x)\cot x=\sin x\cdot(\tan x\cot x)=\sin x,$$ since $\tan x\cot x=1$. For all $x$ where the original expression is defined, this simplification is valid. The choice $\sin x\tan x$ is a tempting intermediate, but it is what $\sec x-\cos x$ alone simplifies to—you still have the $\cot x$ factor to use. The choice $\cos x$ often comes from incorrectly reducing $\sec x-\cos x$ to just $\sin x$, then using $\sin x\cot x=\cos x$; however, the missing $\tan x$ factor is exactly what cancels with $\cot x$. The choice $\sec x$ would mean dropping the $\cot x$ factor entirely or treating it as $1$, but $\cot x$ 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.
3. If $\frac{\sec^2 x-1}{\csc^2 x-1}=k\tan^4 x$ for all $x$ in the domain, what is $k$?
   Answer: $1$ Use $\sec^2 x-1=\tan^2 x$ and $\csc^2 x-1=\cot^2 x$. The quotient is $\frac{\tan^2 x}{\cot^2 x}=\tan^2 x\cdot\tan^2 x=\tan^4 x$, so $k=1$. The other choices come from sign errors or from treating the quotient as $2\tan^4 x$ or $4\tan^4 x$.
4. Given $\sin \theta=\frac{3}{5}$ with $\theta$ in Quadrant II, evaluate $\tan \theta\cot \theta+\cos^2 \theta$.
   Answer: $\frac{41}{25}$ Whenever you see a trig expression involving identities, simplify the structure before plugging in numbers. Here the key simplification is $\tan\theta\cot\theta=1$, since $\cot\theta=\frac{1}{\tan\theta}$. So the first term contributes exactly $1$, regardless of the quadrant. The remaining part uses the Pythagorean identity: $\sin^2\theta+\cos^2\theta=1$. Given $\sin\theta=\frac{3}{5}$, you get $\cos^2\theta=1-\frac{9}{25}=\frac{16}{25}$. Quadrant II tells you $\cos\theta$ is negative, so $\cos\theta=-\frac45$, but $\cos^2\theta$ is still positive $\frac{16}{25}$. Therefore the whole expression is $$1+\frac{16}{25}=\frac{41}{25}.$$ The choice $\frac{16}{25}$ is only the cosine-squared piece; it forgets the $\tan\theta\cot\theta=1$ term. The choice $1$ is only the product identity; it ignores the cosine-squared term entirely. The choice $-\frac{16}{25}$ likely comes from carrying the negative sign of $\cos\theta=-\frac45$ into the square, but a squared value can never be negative. Study tip: first simplify identities like $\tan\theta\cot\theta$, then use Pythagorean identities. Use the quadrant only to determine signs of unsquared trig functions, never to make a square negative.
5. Which expression is equivalent to $(\sec x-\tan x)(\sec x+\tan x)+\frac{1}{\cot^2 x+1}$ on its domain?
   Answer: $1+\sin^2 x$ First, $(\sec x-\tan x)(\sec x+\tan x)=\sec^2 x-\tan^2 x=1$. Also, $\cot^2 x+1=\csc^2 x$, so $\frac{1}{\cot^2 x+1}=\frac{1}{\csc^2 x}=\sin^2 x$. The sum is $1+\sin^2 x$. The expression $1+\cos^2 x$ would come from replacing $\cot^2 x+1$ with $\tan^2 x+1$; $\sec^2 x$ mistakes the reciprocal; and $2$ treats the second term as $1$.
6. Given $\sec \theta-\tan \theta=\frac{1}{3}$ for an acute angle $\theta$, find $\sec \theta+\tan \theta$.
   Answer: $3$ Whenever you see a pairing like $\sec\theta-\tan\theta$ and $\sec\theta+\tan\theta$, think about multiplying conjugates. The key identity is $$(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=\sec^2\theta-\tan^2\theta=1,$$ because $\sec^2\theta=1+\tan^2\theta$. So the product of these two expressions is always exactly $1$. Here you are told the first factor is $\frac{1}{3}$, so $$\frac{1}{3}(\sec\theta+\tan\theta)=1 \quad\Rightarrow\quad \sec\theta+\tan\theta=3.$$ Since $\theta$ is acute, both secant and tangent are positive, so the positive value $3$ is correct. Why the wrong choices miss the mark: choosing $\frac{1}{3}$ simply repeats the given value without using the conjugate relationship. Choosing $\frac{1}{9}$ would be squaring the given value, but the identity calls for multiplication by the conjugate, not squaring. Choosing $9$ comes from inverting the square of the given value, $\left(\frac{1}{3}\right)^{-2}=9$, which incorrectly treats the product as $\frac{1}{9}$ instead of $1$. Study tip: whenever you see $\sec\theta\pm\tan\theta$, immediately recall $$(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1.$$ 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.
7. Which expression is equivalent to $\frac{1}{\sec x+\tan x}+\frac{1}{\sec x-\tan x}$ on its domain?
   Answer: $2\sec x$ Combine over the common denominator $(\sec x+\tan x)(\sec x-\tan x)=\sec^2 x-\tan^2 x=1$. The numerator is $(\sec x-\tan x)+(\sec x+\tan x)=2\sec x$, so the expression equals $2\sec x$. The other choices result from incorrect sign handling or from treating the denominator as something other than 1.
8. A student simplifies $\frac{1-\sin^2 x}{1-\sin x}$ to $1+\sin x$. Which restriction is necessary for the simplification to be valid?
   Answer: It is valid only where $\sin x\ne1$. The simplification factors the numerator as $(1-\sin x)(1+\sin x)$. Canceling $1-\sin x$ requires $1-\sin x\ne0$, so $\sin x\ne1$. At those values the original expression is $\frac00$, while the simplified expression is defined, so the equivalence fails. The cosine and $\sin x=-1$ restrictions are not the cause of the issue.
9. Simplify $\frac{1+\cot^2 x}{1+\tan^2 x}$ for all values of $x$ for which the expression is defined.
   Answer: $\cot^2 x$ Seeing $1+\cot^2 x$ and $1+\tan^2 x$ should immediately trigger the Pythagorean identities: $1+\cot^2 x=\csc^2 x$ and $1+\tan^2 x=\sec^2 x$. These identities hold for every $x$ where the original expression is defined, meaning where both $\sin x$ and $\cos x$ are nonzero. Replacing the numerator and denominator gives $$\frac{\csc^2 x}{\sec^2 x} = \frac{\frac{1}{\sin^2 x}}{\frac{1}{\cos^2 x}} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x.$$ So the simplified form is $\cot^2 x$. The wrong choices each come from a specific misstep. Choosing $\tan^2 x$ is the reciprocal of the correct answer; this happens if you invert the fraction instead of simplifying it. Choosing $\csc^2 x$ treats only the numerator as the answer, ignoring the denominator $\sec^2 x$. Choosing $\sec^2 x$ focuses only on the denominator, or mistakes the denominator alone for the whole simplified expression. For this exam, whenever you see $1+\tan^2 x$ or $1+\cot^2 x$, rewrite them as $\sec^2 x$ and $\csc^2 x$ before doing anything else. Also remember $\cot x = \frac{\cos x}{\sin x}$, not the reciprocal of $\tan x$ in a way that would lead you to invert the final ratio. That one substitution will make this type of problem straightforward.
10. If $\sec\theta=-3$ and $\theta$ lies in Quadrant II, what is $\tan\theta$?
    Answer: $-2\sqrt{2}$ Since $\sec\theta=-3$, $\cos\theta=-\frac13$. Use $1+\tan^2\theta=\sec^2\theta$, so $\tan^2\theta=9-1=8$. In Quadrant II, tangent is negative, so $\tan\theta=-\sqrt{8}=-2\sqrt{2}$. The positive radical has the wrong sign, and the denominator-3 choices confuse tangent with a sine/cosine ratio.
11. Over $0\le x<2\pi$, which set describes all solutions of $\sqrt{1+\tan^2 x}=\sec x$?
    Answer: $0\le x<\frac{\pi}{2}$ or $\frac{3\pi}{2}<x<2\pi$ By the Pythagorean identity, $1+\tan^2 x=\sec^2 x$, so the left side is $\sqrt{\sec^2 x}=|\sec x|$. The equation $|\sec x|=\sec x$ holds exactly when $\sec x\ge 0$, which means $\cos x>0$. This occurs in Quadrants I and IV: $0\le x<\frac{\pi}{2}$ or $\frac{3\pi}{2}<x<2\pi$. The interval where secant is negative is not part of the solution set, and excluding only the undefined points ignores the absolute value.
12. If $\tan\theta+\cot\theta=3$, what is $\tan^2\theta+\cot^2\theta$?
    Answer: $7$ Square the given equation: $(\tan\theta+\cot\theta)^2=9$. Expanding gives $\tan^2\theta+\cot^2\theta+2\tan\theta\cot\theta=9$. Since $\tan\theta\cot\theta=1$, $\tan^2\theta+\cot^2\theta+2=9$, so $\tan^2\theta+\cot^2\theta=7$. The value $9$ omits the product term; the values $5$ and $11$ use the product term with an incorrect sign or coefficient.
13. If $\sin\theta+\cos\theta=\frac12$, what is $\sin\theta\cos\theta$?
    Answer: $-\frac38$ Square both sides: $(\sin\theta+\cos\theta)^2=\frac14$. Expanding gives $\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=\frac14$, so $1+2\sin\theta\cos\theta=\frac14$. Thus $2\sin\theta\cos\theta=-\frac34$, and $\sin\theta\cos\theta=-\frac38$. The value $-\frac34$ comes from forgetting to divide by 2; the positive values have the wrong sign.
14. Which statement is NOT an identity for all values of $x$ for which both sides are defined?
    Answer: $\frac{\sin^2 x}{1-\cos x}=1-\cos x$ Since $\sec x/\tan x = (1/\cos x)/(\sin x/\cos x)=1/\sin x=\csc x$, that equation is true wherever both sides are defined. Since $\cos^2 x=(1-\sin x)(1+\sin x)$, $\frac{\cos^2 x}{1-\sin x}=1+\sin x$ and $\frac{\cos^2 x}{1+\sin x}=1-\sin x$ wherever the denominators are nonzero. However, $\sin^2 x=(1-\cos x)(1+\cos x)$, so $\frac{\sin^2 x}{1-\cos x}=1+\cos x$, not $1-\cos x$. Thus the false statement is $\frac{\sin^2 x}{1-\cos x}=1-\cos x$.
15. Which expression is NOT equivalent to $\sin^2 x$ for all values of $x$ for which it is defined?
    Answer: $\frac{1}{\sec^2 x}$ When you see a question asking whether an expression is equivalent to $\sin^2 x$, your first move should be to rewrite everything in terms of sine and cosine. That instantly makes most equivalences clear. The expression $1-\cos^2 x$ is just a rearrangement of the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, so it is exactly $\sin^2 x$. For $\frac{\tan^2 x}{\sec^2 x}$, rewrite it as $\frac{\sin^2 x/\cos^2 x}{1/\cos^2 x}$, which simplifies to $\sin^2 x$. Similarly, $\frac{1}{\csc^2 x}$ is $\sin^2 x$, because $\csc x = \frac{1}{\sin x}$, so its reciprocal squared is $\sin^2 x$. The trap is $\frac{1}{\sec^2 x}$. Since $\sec x = \frac{1}{\cos x}$, this expression equals $\cos^2 x$, not $\sin^2 x$. 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 $\tan$, $\cot$, $\sec$, and $\csc$ into $\sin$ and $\cos$ 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.
16. Which expression is equivalent to $\frac{\sin \theta}{1+\cos \theta}$ for all values of the variable for which the expression is defined?
    Answer: $\frac{1-\cos \theta}{\sin \theta}$ Whenever you see a single trig fraction, try multiplying by a conjugate. Since the denominator is $1+\cos\theta$, multiply the numerator and denominator by $1-\cos\theta$: $$\frac{\sin\theta}{1+\cos\theta}\cdot \frac{1-\cos\theta}{1-\cos\theta} = \frac{\sin\theta(1-\cos\theta)}{1-\cos^2\theta} = \frac{\sin\theta(1-\cos\theta)}{\sin^2\theta} = \frac{1-\cos\theta}{\sin\theta}.$$ The key step uses the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$. So for values where these expressions are defined, $\frac{1-\cos\theta}{\sin\theta}$ is the equivalent form. The other choices miss this idea. $\frac{1+\cos\theta}{\sin\theta}$ is what you get if you multiply by $1+\cos\theta$ instead, and it is actually paired with $\frac{\sin\theta}{1-\cos\theta}$; those two are equivalent to each other, not to the original. $\frac{1-\sin\theta}{\cos\theta}$ comes from incorrectly replacing parts of $\sin^2+\cos^2=1$, but $1-\sin\theta$ is not a useful identity here. Study tip: for a denominator like $1+\cos\theta$, multiply by its conjugate $1-\cos\theta$. This turns the denominator into $\sin^2\theta$ and usually lets you cancel to a simpler form.
17. Which expression is NOT equivalent to $\sec x\csc x$ on its domain?
    Answer: $\frac{\cot x}{\sin^2 x}$ The sum $\tan x+\cot x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{1}{\sin x\cos x}=\sec x\csc x$. Also, $\frac{\sec^2 x}{\tan x}=\frac{1}{\cos^2 x}\cdot\frac{\cos x}{\sin x}=\sec x\csc x$, and $\frac{\csc^2 x}{\cot x}=\frac{1}{\sin^2 x}\cdot\frac{\sin x}{\cos x}=\sec x\csc x$. The expression with cotangent in the numerator, $\frac{\cot x}{\sin^2 x}=\frac{\cos x}{\sin^3 x}$, is not $\sec x\csc x$.
18. Which expression is equivalent to $\frac{1}{1+\sin x}+\frac{1}{1-\sin x}$ on its domain?
    Answer: $2\sec^2 x$ Combine over the common denominator $(1+\sin x)(1-\sin x)=1-\sin^2 x=\cos^2 x$. The numerator is $(1-\sin x)+(1+\sin x)=2$, so the expression is $\frac{2}{\cos^2 x}=2\sec^2 x$. The cosecant version would require $1-\sin^2 x=\sin^2 x$; the cosine form and the Pythagorean form are not reciprocal forms.
19. Which equation is NOT an identity for all values of $x$ for which both sides are defined?
    Answer: $\frac{\cos x}{1-\sin x}=\frac{1-\sin x}{\cos x}$ 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 $\frac{\tan x}{\sec x}$: since $\tan x=\frac{\sin x}{\cos x}$ and $\sec x=\frac{1}{\cos x}$, this becomes $\sin x$, which equals $\frac{1}{\csc x}$. True. Similarly, $\frac{\cot x}{\csc x}=\frac{\cos x/\sin x}{1/\sin x}=\cos x$, which equals $\frac{1}{\sec x}$. True. Now look at $\frac{\cos x}{1-\sin x}=\frac{1-\sin x}{\cos x}$. Cross-multiplying gives $\cos^2 x=(1-\sin x)^2$. Expanding the right side gives $1-2\sin x+\sin^2 x$, so the equation would require $2\sin x(\sin x-1)=0$. That only works for special angles, not all values, so this is the equation that is NOT an identity. Compare with $\frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x}$. Cross-multiplying gives $\cos^2 x=(1-\sin x)(1+\sin x)=1-\sin^2 x=\cos^2 x$, which is true wherever both sides are defined. This is the standard conjugate identity. Study tip: when you see $1-\sin x$ in a denominator, try multiplying by the conjugate $1+\sin x$. And when verifying identities, cross-multiplying quickly reveals whether two rational trig expressions are genuinely equal.
20. Which expression is equivalent to $\frac{\sec x-\cos x}{\tan x}+\frac{\csc x-\sin x}{\cot x}$ on its domain?
    Answer: $\sin x+\cos x$ Rewrite $\sec x-\cos x=\frac{1-\cos^2 x}{\cos x}=\frac{\sin^2 x}{\cos x}$. Dividing by $\tan x=\frac{\sin x}{\cos x}$ gives $\sin x$. Similarly, $\csc x-\sin x=\frac{\cos^2 x}{\sin x}$, and dividing by $\cot x=\frac{\cos x}{\sin x}$ gives $\cos x$. The sum is $\sin x+\cos x$. The other choices come from incorrect division or from adding the reciprocal forms instead of simplifying.

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