Exercises — Reciprocal identities — cosec, sec, cot in terms of sin, cos, tan

What we do: replace each reciprocal function by "$1$ over its base function," then plug in the known base value.

$\sin 90^{\circ }=1$, so $\csc 90^{\circ }=\frac{1}{\sin 90^{\circ }}=\frac{1}{1}=1.$ $\cos 0^{\circ }=1$, so $\sec 0^{\circ }=\frac{1}{\cos 0^{\circ }}=\frac{1}{1}=1.$ $\tan 45^{\circ }=1$ (a 45° right triangle has equal legs, so rise = run), so $\cot 45^{\circ }=\frac{1}{\tan 45^{\circ }}=\frac{1}{1}=1.$

Answers: $\csc 90^{\circ }=1$, $\sec 0^{\circ }=1$, $\cot 45^{\circ }=1$.

What we do: $\sec \theta =\frac{1}{\cos \theta }$ blows up exactly where its denominator $\cos \theta =0$, because dividing by zero has no value.

$\cos \theta =0$ at $\theta =90^{\circ }$ and $\theta =270^{\circ }$.

Answer: $\sec \theta$ is undefined at $\theta =90^{\circ }$ and $\theta =270^{\circ }$. There $\cos \theta =0$, and $\frac{1}{0}$ is undefined — the graph has a vertical asymptote, not a value.

What we do: flip each base value, then add. From the standard triangles: $\sin 30^{\circ }=\frac{1}{2}$ and $\tan 60^{\circ }=\sqrt{3}$.

$\csc 30^{\circ }=\frac{1}{\sin 30^{\circ }}=\frac{1}{1/2}=2,$ $\cot 60^{\circ }=\frac{1}{\tan 60^{\circ }}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}.$

Why rationalise? $\frac{1}{\sqrt{3}}$ and $\frac{\sqrt{3}}{3}$ are equal; the second is the tidy standard form.

$\csc 30^{\circ }+\cot 60^{\circ }=2+\frac{\sqrt{3}}{3}.$

Answer: $2+\frac{\sqrt{3}}{3}\approx 2.577$.

What we do: rewrite everything in $\sin$ and $\cos$ — the "universal language" — then cancel.

$\frac{\csc \theta }{\cot \theta }=\frac{\frac{1}{\sin \theta }}{\frac{\cos \theta }{\sin \theta }}.$

Why this move: $\csc \theta =\frac{1}{\sin \theta }$ and $\cot \theta =\frac{\cos \theta }{\sin \theta }$ share the same denominator $\sin \theta$, which will cancel cleanly.

Dividing by a fraction means multiplying by its reciprocal: $=\frac{1}{\sin \theta }\times \frac{\sin \theta }{\cos \theta }=\frac{1}{\cos \theta }=\sec \theta .$

Answer: $\sec \theta$.

What we do: find the missing side with Pythagoras, then attach the correct sign for the quadrant, then flip.

Step 1 — the easy flip. $\csc \theta =\frac{1}{\sin \theta }=\frac{1}{-5/13}=-\frac{13}{5}$. (Sign carries straight through: negative sine $\Rightarrow$ negative cosecant.)

Step 2 — find $\cos \theta$ via the Pythagorean identity (see 2.4.05-Pythagorean-identities): ${\sin }^2\theta +{\cos }^2\theta =1\Rightarrow {\cos }^2\theta =1-{\left(-\frac{5}{13}\right)}^2=1-\frac{25}{169}=\frac{144}{169}.$ $\cos \theta =\pm \frac{12}{13}.$

Step 3 — pick the sign. Look at the figure: in the third quadrant (bottom-left) both $x$ and $y$ are negative, so $\cos \theta$ (which tracks $x$) is negative: $\cos \theta =-\frac{12}{13}.$

Step 4 — flip and divide. $\sec \theta =\frac{1}{\cos \theta }=-\frac{13}{12},$ $\cot \theta =\frac{\cos \theta }{\sin \theta }=\frac{-12/13}{-5/13}=\frac{12}{5}.$ Notice $\cot \theta$ is positive: a negative divided by a negative. This matches the ASTC rule — in Quadrant 3, tangent and its flip cotangent are the positive ones.

Answers: $\csc \theta =-\frac{13}{5}$, $\cos \theta =-\frac{12}{13}$, $\sec \theta =-\frac{13}{12}$, $\cot \theta =\frac{12}{5}$.

What we do: use $\cot \theta =\frac{\cos \theta }{\sin \theta }$ and ask two separate questions — when does the denominator vanish (undefined), and when does the numerator vanish (value zero)?

Undefined: when $\sin \theta =0$, i.e. $\theta =0^{\circ }$ and $\theta =180^{\circ }$ (and $360^{\circ }$, outside the half-open range). Dividing by $0$ has no value.

Equal to zero: when the numerator $\cos \theta =0$ but the denominator is fine, i.e. $\theta =90^{\circ }$ and $\theta =270^{\circ }$. There $\cot \theta =\frac{0}{\pm 1}=0$.

Answers: undefined at $\theta =0^{\circ }$ and $180^{\circ }$; equal to zero at $\theta =90^{\circ }$ and $270^{\circ }$.

Strategy: the parent showed $1+{\cot }^2\theta ={\csc }^2\theta$ by dividing ${\sin }^2+{\cos }^2=1$ by ${\sin }^2\theta$. To land on $\sec$ and $\tan$, divide by ${\cos }^2\theta$ instead.

Step 1 — divide by ${\cos }^2\theta$: $\frac{{\sin }^2\theta }{{\cos }^2\theta }+\frac{{\cos }^2\theta }{{\cos }^2\theta }=\frac{1}{{\cos }^2\theta }.$ Why: we want ${\sec }^2\theta =\frac{1}{{\cos }^2\theta }$ on the right, so ${\cos }^2\theta$ must be the divisor.

Step 2 — name each piece: $\frac{{\sin }^2\theta }{{\cos }^2\theta }={\tan }^2\theta$, the middle term is $1$, and the right is ${\sec }^2\theta$: ${\tan }^2\theta +1={\sec }^2\theta .$

Step 3 — rearrange: ${\sec }^2\theta -{\tan }^2\theta =1.■$

What we do: recognise the left side as a fraction begging for the "multiply by the conjugate" trick, and use the identity from 4.1 written for cosec/cot.

Step 1 — the tool we need. From Exercise 3-style Pythagorean division we also have ${\csc }^2\theta -{\cot }^2\theta =1$ (divide ${\sin }^2+{\cos }^2=1$ by ${\sin }^2\theta$: $1+{\cot }^2={\csc }^2$, then move ${\cot }^2$ over). This factors as a difference of squares: $(\csc \theta -\cot \theta )(\csc \theta +\cot \theta )=1.$

Why this tool: the left-hand denominator is $\csc \theta -\cot \theta$, and this identity says that quantity times $(\csc \theta +\cot \theta )$ gives $1$ — exactly a reciprocal relationship.

Step 2 — divide both sides by $(\csc \theta -\cot \theta )$ (allowed since it is non-zero where the expression is defined): $\csc \theta +\cot \theta =\frac{1}{\csc \theta -\cot \theta }.■$

Sanity check at $\theta =30^{\circ }$: $\csc 30^{\circ }=2$, $\cot 30^{\circ }=\sqrt{3}$. LHS $=\frac{1}{2-\sqrt{3}}\approx 3.732$; RHS $=2+\sqrt{3}\approx 3.732$. ✓

What we do: read the reciprocal off the sine graph point by point.

$\sin \theta$ lives entirely in $[-1,1]$. Flipping a number in $[-1,1]$:

• When $\sin \theta =1$, $\csc \theta =1$ (a smallest positive peak).
• As $\sin \theta$ shrinks toward $0^{+}$, $\frac{1}{\sin \theta }$ grows without bound — the orange curve shoots up to $+\infty$ (a vertical asymptote at each zero of sine).
• There is no input whose reciprocal lands strictly between $0$ and $1$, because that would need $\sin \theta >1$, which never happens.

So $\csc \theta \ge 1$ or $\csc \theta \le -1$: it avoids the whole strip $(-1,1)$. In the figure, look at how the orange $\csc$ branches hug the line $y=1$ from above and $y=-1$ from below and never enter the shaded middle band.

Answer: because $|\sin \theta |\le 1$, its reciprocal satisfies $|\csc \theta |\ge 1$; the band $(-1,1)$ is unreachable.

What we do: substitute $\theta =\frac{\pi }{2}$ (radians) into the derivative formula.

At $\theta =\frac{\pi }{2}$: $\csc \frac{\pi }{2}=\frac{1}{\sin (\pi /2)}=\frac{1}{1}=1$ and $\cot \frac{\pi }{2}=\frac{\cos (\pi /2)}{\sin (\pi /2)}=\frac{0}{1}=0$. Therefore ${\frac{d}{d\theta }\csc \theta |}_{\pi /2}=-\csc \frac{\pi }{2}\cot \frac{\pi }{2}=-(1)(0)=0.$

Geometric meaning: slope $0$ means the tangent line to the $\csc$ curve is horizontal at $\theta =\frac{\pi }{2}$ — the curve is momentarily flat. This matches the picture in Exercise 5.1: $\csc$ reaches its local minimum value $1$ exactly where $\sin$ reaches its maximum $1$, and at a turning point the tangent is always horizontal.

Answer: slope $=0$; the point $\left(\frac{\pi }{2},1\right)$ is a local minimum (turning point) of $\csc \theta$.

What we do: turn the reciprocal equation into a cosine equation, then solve it (technique from 2.5.02-Trigonometric-equations-basic).

$\sec \theta =2⟹\frac{1}{\cos \theta }=2⟹\cos \theta =\frac{1}{2}.$

Why flip first: we know cosine's values and quadrants far better than secant's; converting keeps us on familiar ground.

$\cos \theta =\frac{1}{2}$ has reference angle $60^{\circ }$. Cosine is positive in Q1 and Q4: $\theta =60^{\circ }\text{and}\theta =360^{\circ }-60^{\circ }=300^{\circ }.$

Answer: $\theta =60^{\circ },300^{\circ }$.

$\csc 90^{\circ }$ ::: $1$ $\sec 0^{\circ }$ ::: $1$ Where is $\cot \theta$ undefined on $[0^{\circ },360^{\circ })$? ::: at $0^{\circ }$ and $180^{\circ }$ (where $\sin \theta =0$) If $\sec \theta =2$ then $\cos \theta =$ ::: $\frac{1}{2}$ Slope of $\csc \theta$ at $\theta =\frac{\pi }{2}$ (radians) ::: $0$ ${\sec }^2\theta -{\tan }^2\theta =$ ::: $1$