Trigonometry

For complementary angles, $\theta$ and $90^\circ-\theta$:

The 'co-' in cosine literally means 'complement's sine'. Any ratio of one angle equals the co-ratio of its partner.

Check at $\theta=35^\circ$: $\sin35^\circ=\cos55^\circ$, and $\tan35^\circ\tan55^\circ=1$ because the tangents are reciprocal partners. One complement, three identities confirmed on the spot.

$\sec55^\circ=\dfrac{1}{\cos55^\circ}=\dfrac{1}{\sin35^\circ}$, so the first term is $1$. The second term is $1$ the same way. Total: $2$.

Convert everything to one angle, then watch the expression fold.

Another shape: $\dfrac{\cos25^\circ}{\sec65^\circ}+\dfrac{\sin25^\circ}{\cosec65^\circ}$. Since $\cos65^\circ=\sin25^\circ$, the first fraction is $\cos25^\circ\sin25^\circ$; the second equals the same product. The sum is $2\sin25^\circ\cos25^\circ=\sin50^\circ$.

$\tan5^\circ\tan85^\circ=1$ because the angles are complementary. So:

Chains pair off from the ends; the middle $45^\circ$ term is $1$ on its own. Four-factor chains work identically: $\tan10^\circ\tan20^\circ\tan70^\circ\tan80^\circ=1$, since $10+80=90$ and $20+70=90$.

$\cosec68^\circ-\sec22^\circ$: since $68^\circ+22^\circ=90^\circ$, $\cosec68^\circ=\sec22^\circ$. The difference is exactly $0$.

Sums of twins double instead of vanishing: $\sec72^\circ+\cosec18^\circ=2\cosec18^\circ$. And a long chain like $\tan1^\circ\tan2^\circ\cdots\tan89^\circ$ pairs into $44$ ones around the middle $\tan45^\circ=1$, so the whole product is $1$.

$\sin3A=\cos(A-10^\circ)$: the partners rule needs $3A=90^\circ-(A-10^\circ)$, so $4A=100^\circ$ and $A=25^\circ$. Check: $\sin75^\circ=\cos15^\circ$.

Same layout with tangents: $\tan2A=\cot(3A-10^\circ)$ gives $2A+3A-10^\circ=90^\circ$, so $A=20^\circ$. A negative argument also resolves: $\sin(2A-10^\circ)=\cos(3A+20^\circ)$ gives $5A+10^\circ=90^\circ$, so $A=16^\circ$.