14.
\(\dfrac{1-cosa}{sina}=\dfrac{sina\left(1-cosa\right)}{sin^2a}=\dfrac{sina\left(1-cosa\right)}{1-cos^2a}=\dfrac{sin\left(1-cosa\right)}{\left(1-cosa\right)\left(1+cosa\right)}=\dfrac{sina}{1+cosa}\)
Câu b đề bài sai, đẳng thức đúng phải là: \(1+tan^2a=\dfrac{1}{cos^2a}\)
\(1+tan^2a=1+\dfrac{sin^2a}{cos^2a}=\dfrac{sin^2a+cos^2a}{cos^2a}=\dfrac{1}{cos^2a}\)
\(tan^2a-sin^2a=\dfrac{sin^2a}{cos^2a}-sin^2a=\dfrac{sin^2a}{cos^2a}\left(1-cos^2a\right)=\dfrac{sin^2a}{cos^2a}.sin^2a=tan^2a.sin^2a\)
\(\dfrac{sin^4a-cos^4a}{sina+cosa}=\dfrac{\left(sin^2a+cos^2a\right)\left(sin^2a-cos^2a\right)}{sina+cosa}=\dfrac{sin^2a-cos^2a}{sina+cosa}=\dfrac{\left(sina+cosa\right)\left(sina-cosa\right)}{sina+cosa}\)
\(=sina-cosa\)
13.
b. Chia cả tử và mẫu cho sinB:
\(N=\dfrac{\dfrac{4cosB}{sinB}+\dfrac{2sinB}{sinB}}{\dfrac{cossB}{sinB}-\dfrac{3sinB}{sinB}}=\dfrac{4cotB+2}{cotB-3}=\dfrac{4.\dfrac{3}{2}+2}{\dfrac{3}{2}-3}=-\dfrac{16}{3}\)
c. Chia cả tử và mẫu cho \(cos^3B\)
\(M=\dfrac{\dfrac{sin^3B}{cos^3B}-\dfrac{cos^3B}{cos^3B}}{\dfrac{sin^3B}{cos^3B}+\dfrac{cos^3B}{cos^3B}}=\dfrac{tan^3B-1}{tan^3B+1}=\dfrac{3^3-1}{3^3+1}=\dfrac{13}{14}\)
