Question

Prove that : (1-3tan²θ)(3-4sin²θ) = (3-tan²θ)(4cos²θ-3)

Answer 1

Answer:

Step-by-step explanation:

tan(th)=sin(th)/cos(th) so rewrite as

(3–4*sin(th)^2)*(1–3*sin(th)^2/cos(th)^2)=(3-sin(th)^2/cos(th)^2)*(4*cos(th)^2–3) expand out to get

3–4*sin(th)^2–9*sin(th)^2/cos(th)^2+12*sin(th)^4/cos(th)^2=

12*cos(th)^2–9–4*sin(th)^2+3*sin(th)^2/cos(th)^2 subtract both sides to get

12–12*sin(th)^2/cos(th)^2+12*sin(th)^4/cos(th)^2–12*cos(th)^2=0. Divide by 12 and replace sin(th)^2=1-cos(th)^2 to get

1-(1-cos(th)^2)/cos(th)^2+(1-cos(th)^2)^2/cos(th)^2-cos(th)^2=0. Multiply by cos(th)^2 to get

cos(th)^2-(1-cos(th)^2)+(1-cos(th)^2)^2-cos(th)^4=0, expand out

cos(th)^2–1+cos(th)^2+1–2*cos(th)^2+cos(th)^4-cos(th)^4=0 so 0=0 as required.

(th)-theta