Sine, Cosine, and Tangent: Practice

The opposite and adjacent legs have ratio $2:1$. Choose representative lengths $2$ and $1$; Pythagoras gives hypotenuse $\sqrt5$.

Rationalizing gives $\sqrt5/5$. The value $2\sqrt5/5$ is sine, $1/2$ reverses tangent, and $\sqrt5$ reverses cosine.

Both ratios have the same positive hypotenuse denominator.

Multiply by that denominator to obtain equal legs. A leg cannot equal the hypotenuse of a nondegenerate right triangle.

Use the complementary-angle relations in the two products.

The expression becomes the sum of squared sine and cosine.

It is not the unsquared sum. Both products are positive, and the identity fixes their sum at $1$.

Multiply both numerators and both denominators.

Dividing gives $\sqrt5/2$, dropping a denominator factor gives $2\sqrt5/3$, and replacing $\sqrt5$ with $5$ gives $10/9$. The supplied squared values sum to $1$, so they are consistent.

Expand the square, retaining its cross term.

The sum of the two squares is $1$, so the result is $1+2\sin\theta\cos\theta$. The cross term is positive for acute angles. Omitting it or dropping its cosine factor changes the value.

The selected angles are at different corresponding positions. For $\theta$, the opposite leg is $9$ and the hypotenuse is $41$.

For $\phi$ at $E$, the adjacent leg is $EF$, so

Reducing gives the same $9/41$. In contrast, $\sin\phi$ and $\cos\theta$ are $40/41$, and the two tangents are $9/40$ and $40/9$.

Dividing the longer positive leg by the shorter gives a ratio above $1$.

Sine and cosine use the same positive hypotenuse denominator, so their order follows their leg numerators.

Reversing either comparison contradicts the stated leg ordering.

Square the quotient identity.

Use the squared-value identity to replace the denominator.

This gives the stated expression. The denominator is positive for acute angles. Adding instead of subtracting, reversing the quotient, or failing to square its numerator gives the wrong expressions.

The sine and cosine records are consistent because their squares add to $1$. Tangent must be their quotient.

Therefore its exact value is $0.75$. The product is $0.48$, reversing the quotient gives $4/3$, and the squared-value sum is $1$, which is a different quantity.

A hypotenuse belongs to a right triangle and lies opposite its $90^\circ$ angle. This triangle has no right angle.

The acute angle $50^\circ$ can be represented in a separate right triangle, but the three given non-right-triangle sides cannot simply be assigned the right-triangle side-ratio definitions.

The given ratios determine shape but leave the scale free. If the hypotenuse is $13k$, the two ratios require opposite leg $5k$ and adjacent leg $12k$.

Every positive $k$ preserves both ratios and gives a right triangle. Fixing only the hypotenuse prevents arbitrary scaling; making $5k$ the hypotenuse would make it shorter than a leg.

Cosine is positive, so dividing the given inequality by it gives

That quotient is tangent. Sine remains strictly below $1$ because it is a leg divided by the longer hypotenuse. Thus only tangent exceeds $1$.