Cosine over Cosine of Complement plus Sine over Sine of Complement

Theorem

$\dfrac {\cos x} {\map \cos {90 \degrees - x} } + \dfrac {\sin x} {\map \sin {90 \degrees - x} } = \csc x \sec x$

Proof

\(\ds \dfrac {\cos x} {\map \cos {90 \degrees - x} } + \dfrac {\sin x} {\map \sin {90 \degrees - x} }\)

\(=\)

\(\ds \dfrac {\cos x} {\sin x} + \dfrac {\sin x} {\cos x}\)

Cosine of Complement equals Sine, Sine of Complement equals Cosine

\(\ds \)

\(=\)

\(\ds \cot x + \tan x\)

Cotangent is Cosine divided by Sine, Tangent is Sine divided by Cosine

\(\ds \)

\(=\)

\(\ds \csc x \sec x\)

Sum of Tangent and Cotangent

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Exercise $\text {XXXI}$: $3.$