Evaluate the double-angle quotient using the given tangent.

01 / Standard#Your turn

\tan\left(\frac{9\pi}4+x\right)=-7,\quad\frac{1-\sin2x}{\cos2x}

Official paper · jm01-2026 · I.12 · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Reduce the tangent angle modulo π.

Hint 2

Factor both numerator and denominator using sin x and cos x.

Worked solution

The supplied angle differs from x+π/4 by two full tangent periods.
$\frac{\sin x+\cos x}{\cos x-\sin x}=-7$
Use the reciprocal quotient; the supplied value excludes the cancelled zero factors.
$\frac{1-\sin2x}{\cos2x}=\frac{(\cos x-\sin x)^2}{(\cos x-\sin x)(\cos x+\sin x)}=-\frac17$

A: −1/7.

Checks and common pitfalls: The requested quotient is the reciprocal, not the same tangent.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The supplied angle differs from x+π/4 by two full tangent periods.
$\frac{\sin x+\cos x}{\cos x-\sin x}=-7$
Use the reciprocal quotient; the supplied value excludes the cancelled zero factors.
$\frac{1-\sin2x}{\cos2x}=\frac{(\cos x-\sin x)^2}{(\cos x-\sin x)(\cos x+\sin x)}=-\frac17$

Think first. Reveal a hint when the class is ready.

2026 JM01