4 tháng 8 2023 lúc 19:50
a: \(1+tan^2a\)
\(=1+\dfrac{sin^2a}{cos^2a}=\dfrac{cos^2a+sin^2a}{cos^2a}=\dfrac{1}{cos^2a}\)
b: \(1+cot^2a=1+\dfrac{cos^2a}{sin^2a}\)
\(=\dfrac{sin^2a+cos^2a}{sin^2a}=\dfrac{1}{sin^2a}\)
c: \(cot^2a-cos^2a=\dfrac{cos^2a}{sin^2a}-cos^2a\)
\(=cos^2a\left(\dfrac{1}{sin^2a}-1\right)\)
\(=cos^2a\cdot\dfrac{1-sin^2a}{sin^2a}=\dfrac{cos^2a}{sin^2a}\cdot cos^2a=cot^2a\cdot cos^2a\)
d: \(\left(1+cosa\right)\left(1-cosa\right)=1-cos^2a=sin^2a\)
=>\(\dfrac{1+cosa}{sina}=\dfrac{sina}{1-cosa}\)
Đúng 1
Bình luận (0)
