An identity holds for every angle where both sides are defined. Quotient identities tan A= sin Acos A A= cos Asin A Require nonzero denominators. The form cot A= 1tan A also requires tan A to exist and be nonzero. Pythagorean identities sin2A+cos2A=1 1+tan2A=sec2A 1+cot2A=cosec2A The last two follow by dividing by cosine squared or sine squared respectively. Use the same angle throughout. **Correct use** | **Not an identity** sin2 3theta+cos2 3theta=1 | sintheta+costheta=1 sec2 theta2=1+tan2 theta2 | sin4theta+cos4theta=1 Proving an identity Start with one side, express reciprocal ratios in sine and cosine, then combine fractions or factorise until the other side is obtained. **Move** | **Working** Express in sin and cos | cot Asin A = cos Asin A x sin A =cos A Pythagorean swap after factorising | (1-cos A)(1+cos A) =1-cos2 A =sin2 A Combine into one fraction | 1cos2 A-1 = 1-cos2 Acos2 A = sin2 Acos2 A =tan2 A Do not assume the identity you are proving.

An identity holds for every angle where both sides are defined.

\(\tan A=\dfrac{\sin A}{\cos A}\qquad\cot A=\dfrac{\cos A}{\sin A}\)

Require nonzero denominators. The form \(\cot A=\dfrac1{\tan A}\) also requires \(\tan A\) to exist and be nonzero.

\(1+\tan^2A=\sec^2A\qquad1+\cot^2A=\operatorname{cosec}^2A\)

The last two follow by dividing by cosine squared or sine squared respectively. Use the same angle throughout.

\(\sec^2\dfrac{\theta}{2}=1+\tan^2\dfrac{\theta}{2}\)

Start with one side, express reciprocal ratios in sine and cosine, then combine fractions or factorise until the other side is obtained.

\(\begin{aligned} &\cot A\sin A \\ &=\dfrac{\cos A}{\sin A}\times\sin A \\ &=\cos A \end{aligned}\)

\(\begin{aligned} &(1-\cos A)(1+\cos A) \\ &=1-\cos^2 A \\ &=\sin^2 A \end{aligned}\)

\(\begin{aligned} &\dfrac{1}{\cos^2 A}-1 \\ &=\dfrac{1-\cos^2 A}{\cos^2 A} \\ &=\dfrac{\sin^2 A}{\cos^2 A} \\ &=\tan^2 A \end{aligned}\)

Reciprocal Ratios and Identities (HL only)

Reciprocal Ratios and Identities (HL only)