The six trigonometric ratios (right-triangle definition). For an acute angle θ in a right triangle, with sides labelled relative to θ as opposite, adjacent, and the hypotenuse:
Only sinθ and cosθ are truly independent — every other ratio is a quotient or reciprocal built from them, which is why rewriting a hard trig expression purely in sin,cos is such a reliable first move.
Exact values at standard angles (0∘,30∘,45∘,60∘,90∘):
θ
0∘
30∘
45∘
60∘
90∘
sinθ
0
21
21
23
1
cosθ
1
23
21
21
0
tanθ
0
31
1
3
undefined
tan90∘,sec90∘ are undefined because cos90∘=0; csc0∘,cot0∘ are undefined because sin0∘=0. Also sin30∘=cos60∘ and sin60∘=cos30∘ — an early instance of the general complementary-angle pattern sinθ=cos(90∘−θ).
What makes an equation an identity. A trigonometric identity is an equation in trigonometric ratios holding for every value of θ in its domain — not merely for some particular angle. secθ=cosθ1 is an identity (true for all θ with cosθ=0); sinθ=21 is not (true only at specific angles like 30∘ or 150∘).
The three fundamental (Pythagorean) identities, obtained from the Pythagorean theorem applied to a right triangle, divided in turn by the square of the hypotenuse, the adjacent side, and the opposite side:
cos2θ+sin2θ=1,sec2θ−tan2θ=1,csc2θ−cot2θ=1.
Here sin2θ means (sinθ)2, and similarly for the other ratios. Each identity holds wherever both sides are defined — e.g. sec2θ−tan2θ=1 says nothing at θ=90∘, where both terms are individually undefined, but this doesn't stop it from being a genuine identity for every θ where it does make sense.
Reciprocal and quotient identities (restating the ratio definitions as identities in their own right): …
Factoring the Pythagorean identity sec2θ−tan2θ=1 as (secθ−tanθ)(secθ+tanθ)=1 produces a second linear relation between secθ and tanθ; combined with the given p, this is a simple pair of simultaneous equations.
Step 1. Get a second relation from the Pythagorean identity.sec2θ−tan2θ=1 factors as (secθ−tanθ)(secθ+tanθ)=1. Since secθ+tanθ=p,
secθ−tanθ=p1.
Step 2. Solve for secθ. Adding secθ+tanθ=p and secθ−tanθ=p1:
2secθ=p+p1=pp2+1⇒secθ=2pp2+1.
Step 3. Solve for tanθ. Subtracting the two equations instead: …