Question bank — Trigonometric ratios in right triangle — sin, cos, tan, cosec, sec, cot

False — equal $\sin \theta$ means the same shape (same angle θ), but one can be a scaled copy of the other by a factor $k$ (multiply every side by $k$); ratios ignore size, as Similar triangles guarantees.

False — $\sin \theta =\text{opposite}/\text{hypotenuse}$ and the hypotenuse is always the longest side, so the numerator can never exceed the denominator; the ratio stays in $[0,1]$.

True — $\tan \theta =\text{opposite}/\text{adjacent}$, and there is no rule forcing opposite to be smaller than adjacent, so it can exceed $1$ (e.g. a steep triangle with $\theta >45°$).

True — $\sec \theta =\text{hypotenuse}/\text{adjacent}$, and the hypotenuse is the longest side, so this ratio can never dip below $1$ (same for $\csc \theta$).

True — the side opposite A is the side adjacent to B (see the shared figure), so ${\text{opp}}_A/\text{hyp}={\text{adj}}_B/\text{hyp}$; since $A+B=90°$, this is the co-function relation $\sin A=\cos (90°-A)$.

False — it holds for every angle because it is just Pythagoras theorem ($o^2+a^2=h^2$) divided through by $h^2$; see Trigonometric identities.

False — they are the same fact written two ways; $1/\tan \theta =1/(\sin \theta /\cos \theta )=\cos \theta /\sin \theta$.

True for acute angles — inside $0°$ to $90°$ each $\tan$ value occurs once, so equal tangents force equal angles (this uniqueness breaks once you allow angles beyond $90°$, where $\tan$ repeats).

The hypotenuse is opposite the right angle, never the adjacent; "adjacent" means the leg forming one arm of the chosen angle θ (the blue side in the shared figure), so it is a leg, not the hypotenuse.

$\csc \theta$ is the reciprocal of sine ($h/o$), while $\cos \theta$ is $a/h$; the sound-alike names trick people, but they measure completely different ratios.

Opposite and adjacent are angle-dependent; you must use the side across from A itself, otherwise you have computed $\cos A$ (which equals $\sin B$), not $\sin A$.

For acute angles it is fine, but the square root discards sign; beyond a right triangle (via the Unit circle definition of trig functions) cosine can be negative, so the plus-only root is not universally valid.

Doubling numerator and denominator leaves the quotient unchanged; $\tan \theta$ depends on the ratio, not the individual magnitudes.

$2^2+3^2=13\ne 4^2=16$, so this is not a right triangle at all; SOH-CAH-TOA only applies once a right angle exists.

Only sine, tangent, and secant increase with θ on $(0°,90°)$; cosine and cotangent decrease, and cosecant decreases too — they move in opposite directions.

Because scaling a triangle by factor $k$ multiplies every side by $k$, and $k$ cancels in any side-to-side ratio; this is the Similar triangles principle at work (see the scaling figure above).

The side across from angle A is the side touching angle B; since the names are defined relative to the angle, changing the reference angle re-labels the two legs.

The reciprocals (cosec, sec, cot) let you avoid dividing by an expression and make many Trigonometric identities and old table-based calculations cleaner; today three suffice but the six survive by convention.

A line rising at angle θ gains $\text{opposite}$ height per $\text{adjacent}$ horizontal run, and slope is exactly rise over run — so $\tan \theta =o/a$ is the slope, which is why Angle of elevation and depression problems lean on it.

On a unit-length arrow at angle θ, the opposite leg is its height and the adjacent leg its width, so $\sin \theta$ and $\cos \theta$ are those parts — the bridge to Components of vectors and the Unit circle definition of trig functions.

If your computed sin and cos don't satisfy ${\sin }^2+{\cos }^2=1$, you've mislabelled a side or mis-divided, because every genuine right triangle forces this relation.

The adjacent side shrinks toward $0$ while opposite stays finite, so $o/a$ grows without bound — $\tan \theta \to \infty$ and $\tan 90°$ is undefined.

The opposite side shrinks toward $0$, so $\cot \theta =a/o\to \infty$; cotangent blows up at the small-angle end just as tangent does at the large-angle end.

Opposite and adjacent are equal, so $\tan 45°=1$ and $\sin 45°=\cos 45°$; this is the balanced, symmetric right triangle.

No — the right angle is already used, and the other two angles must be acute (sum to $90°$), so θ must stay strictly between $0°$ and $90°$ for a genuine triangle.

The triangle has collapsed flat: θ has shrunk to $0°$, $\sin \theta =0$, $\cos \theta =1$, and $\csc \theta$ is undefined because you'd divide by zero.

At $0°$: $\sin =0$, $\cos =1$; at $90°$: $\sin =1$, $\cos =0$ — they trade roles at the endpoints, foreshadowing the full Unit circle definition of trig functions.