Precalculus

Verifying identities

Someone claims that \displaystyle \frac{\sec x + \csc x}{\tan x + \cot x} and \sin x + \cos x are the same function. Checking a few angles cannot settle it. What does an argument look like that covers every angle at once, and how do you show that a claim like this is false?

You need

The Pythagorean identitiesPrecalculus Simplifying rational expressionsGrades 10 to 11 Rationalizing a denominatorAlgebra 1 CounterexampleGeometry +1 more

Where you are 13 in Trigonometric identities

Trigonometry7

Trigonometric identities13

▴ 5 above Simplifying trig expressions Solving trigonometric equations4

Verifying identities

Product to sum and sum to product formulas Sine and tangent of a sum or difference ▾ Show all 13 (3 more)

When you reach for it

An identity claims that two expressions agree at every input where both are defined, the inputs called the common domain in What an identity is. To verify an identity is to prove that claim. A few matching values prove nothing, because checks cover only the inputs checked. The proof has to be a chain of equalities in which every link is an identity you already trust, and then it covers every input of the common domain at once.

Reach for this when an identity is stated and you are asked to show it holds, when a formula in a later lesson is derived from earlier ones, and when you want to know whether a rewrite you found while simplifying really is safe. Reach for its partner, the Counterexample, when you suspect the claim is false: one input where both sides exist and disagree settles it.

The tools are the ones already in hand. The reciprocal and quotient identities turn every function into sine and cosine, the Pythagorean identities trade a square for its partner, and the algebra of Simplifying rational expressions combines and cancels fractions, with \sin x and \cos x in the place of the variable. Throughout, k stands for any integer.

The procedure

The lesson runs over three days. Day 1 sets the rule that makes a proof a proof and does the most common move, rewriting in sine and cosine. Day 2 adds fractions and conjugates. Day 3 turns to claims that are false, and to choosing a strategy before starting.

Day 1: one side, until it becomes the other

Write a chain from one side to the other. Start with one side, often the more complicated one, and rewrite it one step at a time until it is the other side, letter for letter. Each step replaces an expression by an equal one, so the chain reads

\text{left} = A_1 = A_2 = \cdots = \text{right}

and because equality passes along a chain, the left side equals the right side. Nothing is done to “both sides of the equation”, because until the chain is finished there is no equation known to be true to do anything to.

Why not work on both sides at once. It is tempting to treat the claim as an equation, square both sides or cross-multiply, and stop when a true statement appears. That proves nothing, and here is a false claim it “proves”. Suppose someone says

\cos x = \sqrt{1 - \sin^2 x}

for every x. Square both sides: \cos^2 x = 1 - \sin^2 x, which is the Pythagorean identity, true for every x. Yet at x = \pi the left side is \cos \pi = -1 and the right side is \displaystyle \sqrt{1 - 0} = 1. The claim is false.

Figure 1The graphs of \displaystyle y = \sqrt{1 - \sin^2 x} (solid) and y = \cos x agree wherever \cos x \ge 0. Between \displaystyle \frac{\pi}{2} and \displaystyle \frac{3\pi}{2} the cosine is the dashed curve below the axis. Squaring both sides erases exactly this difference.

The argument ran in the wrong direction. It started from the claim and reached something true, but true statements follow from false ones all the time: -1 = 1 is false, and squaring it gives 1 = 1. What a proof needs is the other direction, from known facts to the claim, and squaring cannot be run backward, since a^2 = b^2 says only that a = b or a = -b. The correct statement is \displaystyle \sqrt{1 - \sin^2 x} = \sqrt{\cos^2 x} = |\cos x|, a chain from one side, and it shows exactly where the claim fails: wherever \cos x < 0.

A chain may also be built from each end: rewrite the left side into some expression C, then the right side into the same C. This is valid only because every step in both halves is reversible: each link is an equality between two expressions, true on the common domain, and an equality reads the same in either direction. So the second half, read backward, runs from C to the right side, and the two halves join into one chain from one side to the other, \text{left} = \cdots = C = \cdots = \text{right} (Example 3 does this). A step that cannot be read backward, such as squaring, never belongs in a chain. What is never allowed is an operation applied to both sides of the unproved claim.

Find the common domain before the first step. List where each side fails to exist: \tan x and \sec x need \cos x \neq 0, \cot x and \csc x need \sin x \neq 0, and every denominator must be nonzero. The identity is claimed on what is left, and every step must hold there. Cancelling a factor is allowed only where that factor is not zero, which the domain guarantees.

When unsure, rewrite in sine and cosine. Using \displaystyle \tan x = \frac{\sin x}{\cos x}, \displaystyle \cot x = \frac{\cos x}{\sin x}, \displaystyle \sec x = \frac{1}{\cos x} and \displaystyle \csc x = \frac{1}{\sin x} turns any expression into one with only two functions, and then the Pythagorean identity \sin^2 x + \cos^2 x = 1 is usually the step that finishes. It is the safe default, not always the shortest path: when both sides are built from \sec x and \tan x, the identity 1 + \tan^2 x = \sec^2 x can do the work directly.

Keep the target in view. At each step, ask what the other side looks like. If it has only \sin x, every \cos^2 x should become 1 - \sin^2 x. If it is a single fraction, combine. If it is a product, factor.

Day 2: fractions and conjugates

Combine fractions over a common denominator. At each input, \sin x and \cos x are numbers, so fractions in them add exactly as numerical fractions do. Look at the product of the denominators before multiplying it out: (\sec x - 1)(\sec x + 1) is \sec^2 x - 1, which is \tan^2 x, and (1 - \cos x)(1 + \cos x) is \sin^2 x.

Multiply by a conjugate. A denominator such as 1 - \sin x is half of a Pythagorean pair. Its conjugate is 1 + \sin x, the same two terms with the sign between them flipped, and

(1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x

turns a two-term denominator into a single square. Multiply top and bottom by the conjugate: that multiplies the fraction by 1, so its value does not change, provided the conjugate is not zero on the common domain. This is the move from rationalizing a denominator, where \displaystyle \frac{1}{3 - \sqrt{2}} is multiplied top and bottom by \displaystyle 3 + \sqrt{2} so that the difference of squares clears the root. Here the difference of squares clears the sum. The pairs that work this way are 1 \pm \sin x, 1 \pm \cos x, \sec x \pm \tan x (product \sec^2 x - \tan^2 x = 1) and \csc x \pm \cot x (product \csc^2 x - \cot^2 x = 1).

Day 3: disproving, and choosing a strategy

One counterexample disproves. An identity claims equality at every input of the common domain, so one input a where both sides are defined and disagree shows the claim is false. Nothing more is needed: no explanation of why, and no list of other failures. An input where one side has no value is never a counterexample, because it is outside the common domain.

Choose test values that can show a difference. At 0, \displaystyle \frac{\pi}{2} and \pi one of \sin x and \cos x is zero, and at \displaystyle \frac{\pi}{4} the two are equal. Many different expressions collapse to the same number there, so a false claim can pass a test at those inputs (Example 7). Values such as \displaystyle \frac{\pi}{6} and \displaystyle \frac{\pi}{3}, where \sin x and \cos x differ and neither is zero, are better first tries. A calculator value such as x = 1 works too, when exact values are not needed.

Test, then prove or disprove. Before a long proof, evaluate both sides at one good value. If they disagree, the claim is false and the work is done. If they agree, that is evidence, not proof, so build the chain. A failed attempt at a chain proves nothing either way: a claim is shown false only by a counterexample.

When a chain stalls. Start from the side with more going on, since simplifying is easier than complicating. Rewrite in sine and cosine. Combine fractions; split a fraction whose numerator is a sum. Look for 1 - \sin^2 x, 1 - \cos^2 x, \sec^2 x - 1 and \csc^2 x - 1, each a single square in disguise. Factor, treating \sin x like a variable. Try a conjugate when a denominator is one of the pairs above. And if the chain from one side stalls, start from the other.

Worked examples

Examples 1 to 3 go with Day 1, Examples 4 to 6 with Day 2, and Examples 7 and 8 with Day 3.

Example 1

Everything in sine and cosine

Verify \displaystyle \frac{\sec x + \csc x}{\tan x + \cot x} = \sin x + \cos x.

Solution

Domain. \sec x and \tan x need \cos x \neq 0; \csc x and \cot x need \sin x \neq 0. The denominator is never zero where it exists, since \displaystyle \tan x + \cot x = \frac{1}{\sin x\cos x}, as the chain below shows. The right side exists everywhere, so the common domain is \displaystyle x \neq \frac{k\pi}{2}.

The left side is the complicated one, and it mixes four functions, so rewrite it in sine and cosine and combine the top and the bottom separately:

\begin{aligned} &\frac{\sec x + \csc x}{\tan x + \cot x} \\ &\quad = \frac{\frac{1}{\cos x} + \frac{1}{\sin x}}{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}} \\ &\quad = \frac{\frac{\sin x + \cos x}{\sin x\cos x}}{\frac{\sin^2 x + \cos^2 x}{\sin x\cos x}} \\ &\quad = \frac{\sin x + \cos x}{\sin^2 x + \cos^2 x} \\ &\quad = \sin x + \cos x \end{aligned}

The step after the two combined fractions multiplies top and bottom by \sin x\cos x, which is not zero on the domain, and the last step uses \sin^2 x + \cos^2 x = 1. The chain runs from the left side to the right side, so the identity holds for \displaystyle x \neq \frac{k\pi}{2}.

Example 2

Two Pythagorean trades

Verify (\sec x - \cos x)(\csc x - \sin x) = \sin x\cos x.

Solution

Domain. The left side needs \cos x \neq 0 and \sin x \neq 0, so the common domain is \displaystyle x \neq \frac{k\pi}{2}.

Each factor on the left becomes one fraction, and each numerator is then a Pythagorean square:

\begin{aligned} &(\sec x - \cos x)(\csc x - \sin x) \\ &\quad = \left(\frac{1 - \cos^2 x}{\cos x}\right)\left(\frac{1 - \sin^2 x}{\sin x}\right) \\ &\quad = \frac{\sin^2 x}{\cos x} \cdot \frac{\cos^2 x}{\sin x} \\ &\quad = \frac{\sin^2 x\cos^2 x}{\sin x\cos x} = \sin x\cos x \end{aligned}

The last step cancels \sin x\cos x, which is not zero on the domain. The right side, \sin x\cos x, also exists at the multiples of \displaystyle \frac{\pi}{2}, where it is 0, but the left side does not, so the identity says nothing there.

Example 3

Staying with secant and tangent

Verify \sec^4 x - \sec^2 x = \tan^4 x + \tan^2 x.

Solution

Domain. Both sides need \cos x \neq 0: \displaystyle x \neq \frac{\pi}{2} + k\pi.

Both sides are built from secant and tangent, and 1 + \tan^2 x = \sec^2 x links the two directly. Factor the left side, then trade:

\begin{aligned} \sec^4 x - \sec^2 x &= \sec^2 x\,(\sec^2 x - 1) \\ &= (1 + \tan^2 x)\tan^2 x \\ &= \tan^2 x + \tan^4 x \end{aligned}

which is the right side with its terms in the other order.

Converting to sine and cosine also works, from each end. The left side is \displaystyle \frac{1}{\cos^4 x} - \frac{1}{\cos^2 x} = \frac{1 - \cos^2 x}{\cos^4 x} = \frac{\sin^2 x}{\cos^4 x}. The right side is

\begin{aligned} \frac{\sin^4 x}{\cos^4 x} + \frac{\sin^2 x}{\cos^2 x} &= \frac{\sin^4 x + \sin^2 x\cos^2 x}{\cos^4 x} \\ &= \frac{\sin^2 x\,(\sin^2 x + \cos^2 x)}{\cos^4 x} \end{aligned}

which is \displaystyle \frac{\sin^2 x}{\cos^4 x} again. Both sides equal the same expression, and reading the second computation backward joins the two into one chain from the left side to the right. It is longer, which is why the first route is the better one here.

Example 4

A difference of fractions

Verify \displaystyle \frac{1}{\sec x - 1} - \frac{1}{\sec x + 1} = 2\cot^2 x.

Solution

Domain. \sec x needs \cos x \neq 0. The denominators need \sec x \neq 1 and \sec x \neq -1, that is \cos x \neq \pm 1, which is the same as \sin x \neq 0. The right side also needs \sin x \neq 0. The common domain is \displaystyle x \neq \frac{k\pi}{2}.

The common denominator is the product (\sec x - 1)(\sec x + 1) = \sec^2 x - 1, and that is \tan^2 x:

\begin{aligned} &\frac{1}{\sec x - 1} - \frac{1}{\sec x + 1} \\ &\quad = \frac{(\sec x + 1) - (\sec x - 1)}{(\sec x - 1)(\sec x + 1)} \\ &\quad = \frac{2}{\sec^2 x - 1} = \frac{2}{\tan^2 x} = 2\cot^2 x \end{aligned}

The last step uses \displaystyle \cot x = \frac{1}{\tan x}, valid because \tan x \neq 0 on the domain. No conversion to sine and cosine was needed: the denominators were a Pythagorean pair already.

Example 5

A fraction minus a cotangent

Verify \displaystyle \frac{\sin x}{1 - \cos x} - \cot x = \csc x.

Solution

Domain. The first fraction needs \cos x \neq 1, that is x \neq 2k\pi, and \cot x needs \sin x \neq 0, that is x \neq k\pi. The second contains the first, and \csc x needs the same, so the common domain is x \neq k\pi.

Write \cot x as a fraction and combine over \sin x\,(1 - \cos x):

\begin{aligned} &\frac{\sin x}{1 - \cos x} - \frac{\cos x}{\sin x} \\ &\quad = \frac{\sin^2 x - \cos x\,(1 - \cos x)}{\sin x\,(1 - \cos x)} \\ &\quad = \frac{\sin^2 x + \cos^2 x - \cos x}{\sin x\,(1 - \cos x)} \\ &\quad = \frac{1 - \cos x}{\sin x\,(1 - \cos x)} \\ &\quad = \frac{1}{\sin x} = \csc x \end{aligned}

The numerator was expanded and regrouped so that \sin^2 x + \cos^2 x appears, and then 1 - \cos x cancels, which is allowed because \cos x \neq 1 on the domain. Keeping the denominator factored until the end is what made the cancellation visible.

Example 6

Multiplying by a conjugate

Verify \displaystyle \frac{\cos x}{1 - \sin x} = \frac{1 + \sin x}{\cos x}, and then write the right side as a sum of two functions.

Solution

Domain. The left side needs \sin x \neq 1, so \displaystyle x \neq \frac{\pi}{2} + 2k\pi. The right side needs \cos x \neq 0, so \displaystyle x \neq \frac{\pi}{2} + k\pi. The common domain is the smaller set, \displaystyle x \neq \frac{\pi}{2} + k\pi.

The denominator 1 - \sin x is half of the pair (1 - \sin x)(1 + \sin x) = \cos^2 x, so multiply top and bottom by 1 + \sin x. On the domain \sin x \neq -1, so the factor is not zero and the step multiplies by 1:

\begin{aligned} &\frac{\cos x}{1 - \sin x} \\ &\quad = \frac{\cos x\,(1 + \sin x)}{(1 - \sin x)(1 + \sin x)} \\ &\quad = \frac{\cos x\,(1 + \sin x)}{1 - \sin^2 x} \\ &\quad = \frac{\cos x\,(1 + \sin x)}{\cos^2 x} = \frac{1 + \sin x}{\cos x} \end{aligned}

cancelling the nonzero \cos x. Splitting the numerator,

\frac{1 + \sin x}{\cos x} = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sec x + \tan x

so all three expressions agree for \displaystyle x \neq \frac{\pi}{2} + k\pi. At \displaystyle x = \frac{3\pi}{2} the left side is \displaystyle \frac{0}{2} = 0 while the other two have no value: that input is outside the common domain, and the identity says nothing about it.

Example 7

A test that passes by luck

Mikko remembers the identity as \displaystyle \frac{1 - \cos x}{\sin x} = \frac{\sin x}{1 - \cos x} and checks it at \displaystyle x = \frac{\pi}{2}, where both sides are 1. Is it an identity?

Solution

Both sides need \sin x \neq 0 and \cos x \neq 1, so the common domain is x \neq k\pi. Test \displaystyle x = \frac{\pi}{3}, where \displaystyle \sin x = \frac{\sqrt{3}}{2} and \displaystyle 1 - \cos x = \frac{1}{2}:

\begin{aligned} \frac{1 - \cos\frac{\pi}{3}}{\sin\frac{\pi}{3}} &= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{3} \\ \frac{\sin\frac{\pi}{3}}{1 - \cos\frac{\pi}{3}} &= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \end{aligned}

Both sides are defined and \displaystyle \frac{\sqrt{3}}{3} \neq \sqrt{3}, so it is not an identity.

His test was unlucky, not wrong. The two sides are reciprocals, and a number equals its reciprocal only when it is 1 or -1. On the common domain \displaystyle \left(\frac{1 - \cos x}{\sin x}\right)^2 = 1 means (1 - \cos x)^2 = 1 - \cos^2 x, that is (1 - \cos x)(-2\cos x) = 0, and since \cos x \neq 1 this happens only when \cos x = 0. So the sides agree exactly at \displaystyle x = \frac{\pi}{2} + k\pi, and his one test landed on one of those inputs.

Figure 2The two sides of the claim in Example 7. They meet only at \displaystyle x = \frac{\pi}{2} and \displaystyle x = \frac{3\pi}{2}, so a test at \displaystyle \frac{\pi}{2} finds agreement by bad luck.

The identity he was reaching for is \displaystyle \frac{1 - \cos x}{\sin x} = \frac{\sin x}{1 + \cos x}, which the conjugate 1 + \cos x proves in one line.

Example 8

Test first, then prove or disprove

Niilo and Paavo each propose an identity. Decide whether each is true, and prove it or give a counterexample.

(a) Niilo: \displaystyle \frac{\sin x + \tan x}{1 + \cos x} = \tan x \qquad (b) Paavo: \displaystyle \sec x + \csc x = \frac{1}{\sin x + \cos x}

Solution

(a) Test. At \displaystyle x = \frac{\pi}{3}: \displaystyle \sin x + \tan x = \frac{\sqrt{3}}{2} + \sqrt{3} = \frac{3\sqrt{3}}{2} and \displaystyle 1 + \cos x = \frac{3}{2}, so the left side is \displaystyle \sqrt{3}, which is \displaystyle \tan\frac{\pi}{3}. The test passes, so try to prove it.

Domain. \tan x needs \cos x \neq 0, and the denominator needs \cos x \neq -1. The common domain is \displaystyle x \neq \frac{\pi}{2} + k\pi and x \neq \pi + 2k\pi.

Chain. The numerator has a common factor once \tan x is written as a fraction:

\begin{aligned} \sin x + \frac{\sin x}{\cos x} &= \frac{\sin x\cos x + \sin x}{\cos x} \\ &= \frac{\sin x\,(\cos x + 1)}{\cos x} \end{aligned}

Dividing by 1 + \cos x, which is not zero on the domain, cancels the factor:

\begin{aligned} \frac{\sin x + \tan x}{1 + \cos x} &= \frac{\sin x\,(1 + \cos x)}{\cos x\,(1 + \cos x)} \\ &= \frac{\sin x}{\cos x} = \tan x \end{aligned}

It is an identity on the stated domain.

(b) Test. At \displaystyle x = \frac{\pi}{4} both sides are defined: \displaystyle \sec\frac{\pi}{4} + \csc\frac{\pi}{4} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}, while

\frac{1}{\sin\frac{\pi}{4} + \cos\frac{\pi}{4}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}

Since \displaystyle 2\sqrt{2} \neq \frac{\sqrt{2}}{2}, it is not an identity. A special angle was fine here: it showed a difference, and a difference at any input of the common domain is a counterexample. The claim treats the reciprocal of a sum as the sum of the reciprocals; the true statement is \displaystyle \sec x + \csc x = \frac{\sin x + \cos x}{\sin x\cos x}.

Boundaries

An identity holds on its common domain, not beyond it. A verified identity says nothing at inputs where either side is undefined, and the two sides may still be different functions there (Examples 2 and 6).

Every step must be valid on the whole common domain. Cancelling a factor or multiplying top and bottom by a conjugate needs that factor to be nonzero at every input of the domain. If it can be zero somewhere in the domain, the step fails at those inputs, and they need separate treatment.

Operations on both sides of an unproved claim prove nothing. Squaring loses the sign, so it can turn a false claim into a true one. Steps that can be undone, such as adding the same expression to both sides or multiplying both sides by something nonzero, can be turned into a proof by running them backward, but that has to be done and checked step by step. A chain from one side to the other needs no such repair.

Agreement at test values is not a proof. Any number of matching values leaves the other inputs unchecked; only a chain covers them all. Disagreement at one value of the common domain is a complete disproof.

A square root of a square is an absolute value. \displaystyle \sqrt{\cos^2 x} = |\cos x|, not \cos x. An identity with a root in it usually holds only where the sign is fixed.

Where this goes next

Every identity in the rest of this unit is proved by this method. The sum and difference formulas are derived once and then used as links in chains, and the double angle formulas and power reducing formulas each come from a short chain starting at an earlier formula. Solving trigonometric equations uses the same rewriting with the opposite goal: there the equation is conditional, and the rewriting turns it into one simple enough to solve, with the domain bookkeeping deciding which candidates survive. In calculus, Trigonometric integrals start by rewriting a product of powers with exactly these moves, and a conjugate often clears a denominator first.

Exercises

Day 1: one side, rewriting in sine and cosine

Verify each identity and state its common domain.

\cot x\sec x = \csc x
\cos x\,(\sec x - \cos x) = \sin^2 x
\sin^2 x\sec^2 x + \sin^2 x\csc^2 x = \sec^2 x
\csc^2 x - \cot^2 x\cos^2 x = 1 + \cos^2 x
\tan^2 x - \sin^2 x = \tan^2 x\sin^2 x
\displaystyle \frac{1 + \tan^2 x}{1 + \cot^2 x} = \tan^2 x

Day 2: fractions and conjugates

Verify each identity and state its common domain.

\displaystyle \frac{\sin x}{\csc x - 1} + \frac{\sin x}{\csc x + 1} = 2\tan^2 x
\displaystyle \frac{\sec x}{\sin x} - \frac{\sin x}{\cos x} = \cot x
\displaystyle \frac{\cos x}{1 - \tan x} + \frac{\sin x}{1 - \cot x} = \sin x + \cos x
\displaystyle \frac{\cos x}{1 + \sin x} = \sec x - \tan x
\displaystyle \frac{1}{\sec x - \tan x} = \sec x + \tan x
\displaystyle \frac{1 + \sin x}{1 - \sin x} = (\sec x + \tan x)^2
\displaystyle \frac{\tan x}{\sec x + 1} = \frac{\sec x - 1}{\tan x}

Day 3: counterexamples

Show that each equation is not an identity.

\sec x - \cos x = \sin x. First show that the two sides agree at \displaystyle x = \frac{\pi}{4}.
\tan^2 x + \cot^2 x = 1
\displaystyle \sqrt{\sec^2 x - 1} = \tan x
\displaystyle \frac{\sin^2 x}{1 + \cos x} = 1 + \cos x. Show that \displaystyle x = \frac{\pi}{2} is not a counterexample, then find one.

Day 3: verify or disprove

Decide whether each equation is an identity. Verify it, with its common domain, or give a counterexample.

\sin^3 x + \sin x\cos^2 x = \sin x
\displaystyle \frac{\csc x - \sin x}{\cot x} = \cos x
\displaystyle \frac{\tan x - \sin x}{\sin^3 x} = \frac{\sec x}{1 + \cos x}
(1 + \tan x)^2 = \sec^2 x + \tan x

Reasoning

Paavo “proves” \displaystyle \sin x = \sqrt{1 - \cos^2 x} by squaring both sides and getting the Pythagorean identity. Explain what is wrong, and give a counterexample.
Niilo verifies \tan x + \cot x = \sec x\csc x by rewriting the left side as \displaystyle \frac{1}{\sin x\cos x} and, separately, the right side as \displaystyle \frac{1}{\sin x\cos x}. Is this a proof? Write his work as a single chain from the left side to the right side.
Mikko cross-multiplies the identity of Example 6, \displaystyle \frac{\cos x}{1 - \sin x} = \frac{1 + \sin x}{\cos x}, to get \cos^2 x = (1 - \sin x)(1 + \sin x), which is true. Is his work a proof as it stands? Explain how to turn it into one, and why the same repair is impossible for the squaring argument in the procedure.
Explain why squaring both sides can turn a false claim into a true statement, while adding the same expression to both sides cannot.

Answers

Common domain \displaystyle x \neq \frac{k\pi}{2}. \displaystyle \cot x\sec x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos x}, which is \displaystyle \frac{1}{\sin x} = \csc x.
Common domain \displaystyle x \neq \frac{\pi}{2} + k\pi. \cos x\sec x - \cos^2 x = 1 - \cos^2 x, which is \sin^2 x.
Common domain \displaystyle x \neq \frac{k\pi}{2}. \displaystyle \sin^2 x\sec^2 x = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x and \sin^2 x\csc^2 x = 1, so the left side is \tan^2 x + 1 = \sec^2 x.
Common domain x \neq k\pi. Over \sin^2 x the left side is \displaystyle \frac{1 - \cos^4 x}{\sin^2 x}. The top factors as (1 - \cos^2 x)(1 + \cos^2 x), which is \sin^2 x\,(1 + \cos^2 x). Cancelling \sin^2 x leaves 1 + \cos^2 x.
Common domain \displaystyle x \neq \frac{\pi}{2} + k\pi. Over \cos^2 x the left side is \displaystyle \frac{\sin^2 x\,(1 - \cos^2 x)}{\cos^2 x}. The top is \sin^2 x \cdot \sin^2 x, so this is \displaystyle \frac{\sin^2 x}{\cos^2 x} \cdot \sin^2 x, which is \tan^2 x\sin^2 x.
Common domain \displaystyle x \neq \frac{k\pi}{2}. \displaystyle \frac{1 + \tan^2 x}{1 + \cot^2 x} = \frac{\sec^2 x}{\csc^2 x}, and \displaystyle \frac{\sec^2 x}{\csc^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x.
Common domain \displaystyle x \neq \frac{k\pi}{2} (\csc x \neq \pm 1 means \sin x \neq \pm 1). The common denominator is (\csc x - 1)(\csc x + 1), which is \csc^2 x - 1 = \cot^2 x. The numerator is \sin x\,(\csc x + 1) + \sin x\,(\csc x - 1), which is 2\sin x\csc x = 2. So the sum is \displaystyle \frac{2}{\cot^2 x} = 2\tan^2 x.
Common domain \displaystyle x \neq \frac{k\pi}{2}. Over \sin x\cos x the left side is \displaystyle \frac{1 - \sin^2 x}{\sin x\cos x}, which is \displaystyle \frac{\cos^2 x}{\sin x\cos x} = \frac{\cos x}{\sin x} = \cot x.
Common domain \displaystyle x \neq \frac{k\pi}{2} and \displaystyle x \neq \frac{\pi}{4} + k\pi. Multiply the first fraction top and bottom by \cos x and the second by \sin x: the terms become \displaystyle \frac{\cos^2 x}{\cos x - \sin x} and \displaystyle \frac{\sin^2 x}{\sin x - \cos x}. Their sum is \displaystyle \frac{\cos^2 x - \sin^2 x}{\cos x - \sin x}, which is \cos x + \sin x after cancelling the nonzero \cos x - \sin x.
Common domain \displaystyle x \neq \frac{\pi}{2} + k\pi. Multiply top and bottom by 1 - \sin x, nonzero there. The top becomes \cos x\,(1 - \sin x) and the bottom 1 - \sin^2 x = \cos^2 x. Cancelling \cos x leaves \displaystyle \frac{1 - \sin x}{\cos x}, which is \sec x - \tan x.
Common domain \displaystyle x \neq \frac{\pi}{2} + k\pi. Multiply top and bottom by \sec x + \tan x, which is \displaystyle \frac{1 + \sin x}{\cos x} and nonzero there. The bottom becomes \sec^2 x - \tan^2 x = 1, leaving \sec x + \tan x.
Common domain \displaystyle x \neq \frac{\pi}{2} + k\pi. Start on the right: \displaystyle (\sec x + \tan x)^2 = \frac{(1 + \sin x)^2}{\cos^2 x}. Write \cos^2 x = (1 - \sin x)(1 + \sin x) and cancel the nonzero 1 + \sin x to get \displaystyle \frac{1 + \sin x}{1 - \sin x}.
Common domain \displaystyle x \neq \frac{k\pi}{2}. Multiply top and bottom by \sec x - 1, nonzero because \cos x \neq 1. The bottom becomes \sec^2 x - 1 = \tan^2 x, so the left side is \displaystyle \frac{\tan x\,(\sec x - 1)}{\tan^2 x}, which is \displaystyle \frac{\sec x - 1}{\tan x}.
At \displaystyle \frac{\pi}{4}: \displaystyle \sqrt{2} - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}, which is \displaystyle \sin\frac{\pi}{4}. At \displaystyle x = \frac{\pi}{3}: \displaystyle \sec x - \cos x = 2 - \frac{1}{2} = \frac{3}{2}, but \displaystyle \sin x = \frac{\sqrt{3}}{2}. The identity is \sec x - \cos x = \sin x\tan x.
At \displaystyle x = \frac{\pi}{4}: 1 + 1 = 2 \neq 1.
At \displaystyle x = \frac{3\pi}{4}: \sec^2 x - 1 = 2 - 1 = 1, so the left side is 1, but \tan x = -1. The identity is \displaystyle \sqrt{\sec^2 x - 1} = |\tan x|.
At \displaystyle \frac{\pi}{2} both sides are 1, so it is not a counterexample. At \displaystyle x = \frac{\pi}{3} the left side is \displaystyle \frac{3/4}{3/2} = \frac{1}{2}, but \displaystyle 1 + \cos x = \frac{3}{2}. The identity is \displaystyle \frac{\sin^2 x}{1 + \cos x} = 1 - \cos x.
Identity, for every real x. \sin x\,(\sin^2 x + \cos^2 x) = \sin x.
Identity, for \displaystyle x \neq \frac{k\pi}{2}. \displaystyle \csc x - \sin x = \frac{1 - \sin^2 x}{\sin x}, which is \displaystyle \frac{\cos^2 x}{\sin x}. Dividing by \displaystyle \frac{\cos x}{\sin x} gives \displaystyle \frac{\cos^2 x}{\sin x} \cdot \frac{\sin x}{\cos x} = \cos x.
Identity, for \displaystyle x \neq \frac{k\pi}{2}. \displaystyle \tan x - \sin x = \frac{\sin x\,(1 - \cos x)}{\cos x}, so the left side is \displaystyle \frac{1 - \cos x}{\cos x\sin^2 x}. Write \sin^2 x = (1 - \cos x)(1 + \cos x) and cancel the nonzero 1 - \cos x: the result is \displaystyle \frac{1}{\cos x\,(1 + \cos x)}, which is \displaystyle \frac{\sec x}{1 + \cos x}.
Not an identity. At \displaystyle x = \frac{\pi}{4}: (1 + 1)^2 = 4, but \sec^2 x + \tan x = 2 + 1 = 3. Expanding gives 1 + 2\tan x + \tan^2 x, which is \sec^2 x + 2\tan x.
The argument runs from the claim to a true statement, which proves nothing, and squaring cannot be undone: a^2 = b^2 gives only a = \pm b. Counterexample \displaystyle x = \frac{3\pi}{2}: \sin x = -1, but \displaystyle \sqrt{1 - 0} = 1. The identity is \displaystyle \sqrt{1 - \cos^2 x} = |\sin x|.
Yes. If the left side equals C and the right side equals C, they are equal: each of his steps is an equality, so the right-side half can be read backward. As one chain, for \displaystyle x \neq \frac{k\pi}{2}: \displaystyle \tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x}, which is \displaystyle \frac{1}{\sin x\cos x}, and that is \displaystyle \frac{1}{\cos x} \cdot \frac{1}{\sin x} = \sec x\csc x.
Not as it stands: it runs from the claim to a true statement. But each step can be reversed on the common domain \displaystyle x \neq \frac{\pi}{2} + k\pi: start from \cos^2 x = (1 - \sin x)(1 + \sin x), which is the Pythagorean identity, and divide both sides by \cos x\,(1 - \sin x), which is nonzero there; the result is the claim. Squaring has no such reverse step, because from a^2 = b^2 one cannot get back to a = b.
Adding E to both sides can be undone by subtracting E, so L + E = R + E holds exactly where L = R does; a false claim stays false. Squaring cannot be undone: L^2 = R^2 holds whenever L = -R as well, so a claim false wherever the sides have opposite signs, such as -1 = 1, becomes true.