AHL 3.9—Reciprocal trig ratios and their pythagorean identities. Inverse circular functions

The three reciprocal trigonometric functions are defined as follows: cosecant is the reciprocal of sine, and secant is the reciprocal of cosine.

$\sec \theta =\frac{1}{\cos \theta },\csc \theta =\frac{1}{\sin \theta }$

The option stating $\sec \theta =\frac{1}{\cos \theta }$ and $\csc \theta =\frac{1}{\sin \theta }$ is correct. A common trap is swapping the two — note that sec pairs with cos (both lack a 'co' prefix or both have it, depending on how you remember it).

We need the correct pairing of $\sec \theta$ and $\csc \theta$ with their reciprocal expressions.

The option '$\sec \theta =\frac{1}{\sin \theta }$ and $\csc \theta =\frac{1}{\cos \theta }$' reverses the definitions — this is the most common mistake and is incorrect.

The option '$\sec \theta =\frac{\cos \theta }{\sin \theta }$ and $\csc \theta =\frac{\sin \theta }{\cos \theta }$' describes ratios related to $\cot \theta$ and $\tan \theta$, not $\sec$ and $\csc$ — incorrect.

The option '$\sec \theta =\sin \theta$ and $\csc \theta =\cos \theta$' confuses function names entirely — incorrect.

The remaining option, $\sec \theta =\frac{1}{\cos \theta }$ and $\csc \theta =\frac{1}{\sin \theta }$, is the correct definition.

The Pythagorean identity involving $\csc$ and $\cot$ is: $1+{\cot }^2\alpha ={\csc }^2\alpha$

Subtracting ${\cot }^2\alpha$ from both sides: ${\csc }^2\alpha -{\cot }^2\alpha =1$

The expression simplifies to 1 for all valid values of $\alpha$ (i.e. wherever $\sin \alpha \ne 0$).

We need to evaluate ${\csc }^2\alpha -{\cot }^2\alpha$ as a simplified constant or expression.

If the answer were $0$, that would imply ${\csc }^2\alpha ={\cot }^2\alpha$, i.e. $\frac{1}{{\sin }^2\alpha }=\frac{{\cos }^2\alpha }{{\sin }^2\alpha }$, meaning ${\cos }^2\alpha =1$ always — this is false.

Testing $\alpha =\frac{\pi }{4}$: ${\csc }^2\frac{\pi }{4}=2$ and ${\cot }^2\frac{\pi }{4}=1$, giving $2-1=1\ne 2$ — so 2 is incorrect.

At $\alpha =\frac{\pi }{4}$, ${\sin }^2\frac{\pi }{4}=\frac{1}{2}$, but we computed ${\csc }^2\frac{\pi }{4}-{\cot }^2\frac{\pi }{4}=1\ne \frac{1}{2}$ — so ${\sin }^2\alpha$ is incorrect.

From the identity $1+{\cot }^2\alpha ={\csc }^2\alpha$, it follows directly that ${\csc }^2\alpha -{\cot }^2\alpha =1$.