Exercises — Reciprocal identities, quotient identities

WHAT: Replace each reciprocal name with the flipped fundamental ratio. WHY: By definition $\csc =1/\sin$, $\sec =1/\cos$, $\cot =1/\tan$, and $\tan =\sin /\cos$ so $1/\tan =\cos /\sin$. $\csc \theta =\frac{1}{\sin \theta },\sec \theta =\frac{1}{\cos \theta },\cot \theta =\frac{\cos \theta }{\sin \theta }.$

WHAT: Just flip. WHY: Cosecant is defined as the reciprocal of sine — nothing else needed. $\csc \theta =\frac{1}{\sin \theta }=\frac{1}{5/13}=\frac{13}{5}.$

False. Secant pairs with cosine, not sine (the "co-swap" rule). Correct: $\sec \theta =\frac{1}{\cos \theta }$.

WHAT: Read off $A=x=-8$, $O=y=6$, then $H=\sqrt{(-8{)}^2+6^2}=\sqrt{64+36}=\sqrt{100}=10$. WHY $H$ is always positive: $H$ is a distance from the origin — distances can't be negative. (See figure, Quadrant II.) $\sin \theta =\frac{O}{H}=\frac{6}{10}=\frac{3}{5},\cos \theta =\frac{A}{H}=\frac{-8}{10}=-\frac{4}{5},\tan \theta =\frac{O}{A}=\frac{6}{-8}=-\frac{3}{4}.$ Flip each for the reciprocals: $\csc \theta =\frac{5}{3},\sec \theta =-\frac{5}{4},\cot \theta =-\frac{4}{3}.$

WHAT: Need $\sin$ first, then divide. WHY use Pythagoras: the quotient identity $\tan =\sin /\cos$ needs $\sin$, which we don't have yet — Pythagorean identities supply it. $\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$ (positive because acute). Then $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{4/5}{3/5}=\frac{4}{3}.$

$\cot \theta =\frac{1}{\tan \theta }=\frac{24}{7}$ (flip). $\csc \theta \cdot \sin \theta =\frac{1}{\sin \theta }\cdot \sin \theta =1$ — a reciprocal times its own base is always $1$, whatever $\theta$ is.

WHAT: Convert everything to $\sin ,\cos$ (the universal move). $\frac{\sec \theta }{\csc \theta }=\frac{1/\cos \theta }{1/\sin \theta }=\frac{1}{\cos \theta }\cdot \frac{\sin \theta }{1}=\frac{\sin \theta }{\cos \theta }=\tan \theta .$

$\cot \theta \cdot \sec \theta =\frac{\cos \theta }{\sin \theta }\cdot \frac{1}{\cos \theta }=\frac{1}{\sin \theta }=\csc \theta .$ WHY it collapses: the $\cos \theta$ in $\cot$'s numerator cancels the $\cos \theta$ in $\sec$'s denominator, leaving $1/\sin =\csc$.

$\frac{1+{\cot }^2\theta }{\csc \theta }=\frac{{\csc }^2\theta }{\csc \theta }=\csc \theta .$ WHY: ${\csc }^2/\csc =\csc$ — one factor cancels. (The identity $1+{\cot }^2={\csc }^2$ itself comes from dividing ${\sin }^2+{\cos }^2=1$ by ${\sin }^2$; see Pythagorean identities.)

WHAT: Turn each reciprocal into a division, which becomes a multiplication. $\frac{\sin \theta }{\csc \theta }=\frac{\sin \theta }{1/\sin \theta }=\sin \theta \cdot \sin \theta ={\sin }^2\theta ,$ $\frac{\cos \theta }{\sec \theta }=\frac{\cos \theta }{1/\cos \theta }={\cos }^2\theta .$ Add: ${\sin }^2\theta +{\cos }^2\theta =1$ by the Pythagorean identities. Proved.

WHAT: Convert both to $\sin ,\cos$ and put over a common denominator. $\tan \theta +\cot \theta =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }.$ WHY the common denominator $\sin \theta \cos \theta$: it lets both fractions merge into one; the numerator then becomes ${\sin }^2+{\cos }^2=1$. $=\frac{1}{\sin \theta \cos \theta }.■$

WHAT: The difference of squares factors: ${\sec }^2\theta -{\tan }^2\theta =(\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1$. WHY that identity holds: it is $1+{\tan }^2\theta ={\sec }^2\theta$ rearranged (Pythagoras divided by ${\cos }^2$). $(4)(\sec \theta -\tan \theta )=1⟹\sec \theta -\tan \theta =\frac{1}{4}.$

$A=-1,O=-\sqrt{3},H=\sqrt{(-1{)}^2+(-\sqrt{3}{)}^2}=\sqrt{1+3}=2$. $\sin \theta =\frac{-\sqrt{3}}{2},\cos \theta =\frac{-1}{2},\tan \theta =\frac{-\sqrt{3}}{-1}=\sqrt{3}.$ Reciprocals: $\csc \theta =\frac{-2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3},\sec \theta =-2,\cot \theta =\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}.$ Positive ones: $\tan \theta$ and $\cot \theta$. WHY: in Q3 both $x$ and $y$ are negative, so $\tan =O/A$ is negative÷negative $=$ positive; sine and cosine (and their flips) stay negative. See figure.

WHAT: Convert to $\sin ,\cos$. $\csc \theta =\frac{1}{\sin \theta },\cot \theta +\tan \theta =\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }=\frac{{\cos }^2\theta +{\sin }^2\theta }{\sin \theta \cos \theta }=\frac{1}{\sin \theta \cos \theta }.$ So $\frac{\csc \theta }{\cot \theta +\tan \theta }=\frac{1/\sin \theta }{1/(\sin \theta \cos \theta )}=\frac{1}{\sin \theta }\cdot \sin \theta \cos \theta =\cos \theta .$

Proof. Multiply top and bottom by the conjugate $\sec \theta +\tan \theta$: $\frac{1}{\sec \theta -\tan \theta }\cdot \frac{\sec \theta +\tan \theta }{\sec \theta +\tan \theta }=\frac{\sec \theta +\tan \theta }{{\sec }^2\theta -{\tan }^2\theta }.$ WHY the conjugate: it turns the denominator into a difference of squares, and ${\sec }^2\theta -{\tan }^2\theta =1$ (Pythagoras). So the whole thing $=\sec \theta +\tan \theta$. $■$

Numeric check at $45°$: LHS $=\frac{1}{\sqrt{2}-1}$. Rationalise: $\frac{1}{\sqrt{2}-1}\cdot \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{\sqrt{2}+1}{2-1}=\sqrt{2}+1$. RHS $=\sqrt{2}+1$. ✓