a.

\(tana=\dfrac{sina}{cosa}=\dfrac{1}{15}\Rightarrow sina=\dfrac{cosa}{15}\)

\(\Rightarrow sin2a=2sina.cosa=\dfrac{2cosa}{15}.cosa=\dfrac{2}{15}cos^2a=\dfrac{2}{15}.\dfrac{1}{1+tan^2a}=\dfrac{2}{15}.\dfrac{1}{1+\dfrac{1}{15^2}}=\dfrac{15}{113}\)

b.

\(5^2=\left(3sina+4cosa\right)^2\le\left(3^2+4^2\right)\left(sin^2+cos^2a\right)=25\)

Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\dfrac{sina}{3}=\dfrac{cosa}{4}\\3sina+4cosa=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}sina=\dfrac{3}{5}\\cosa=\dfrac{4}{5}\end{matrix}\right.\)

c.

\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)

\(\Leftrightarrow\dfrac{cos^2a}{sin^2a}+\dfrac{sin^2a}{cos^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)

\(\)\(\Leftrightarrow\dfrac{sin^4a+cos^4a}{sin^2a.cos^2a}+\dfrac{sin^2a+cos^2a}{sin^2a.cos^2a}=7\)

\(\Leftrightarrow\dfrac{\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a}{sin^2a.cos^2a}+\dfrac{1}{sin^2a.cos^2a}=7\)

\(\Leftrightarrow\dfrac{2}{sin^2a.cos^2a}=9\)

\(\Leftrightarrow\dfrac{8}{\left(2sina.cosa\right)^2}=9\)

\(\Leftrightarrow\dfrac{8}{sin^22a}=9\)

\(\Leftrightarrow sin^22a=\dfrac{8}{9}\)

app hay 

2a-ab+b=5 <=> a(2-b) - (2-b) = 3 <=> (a-1)(2-b) = 3

Tới đây phân tích 3 = 1 x 3 = -1 x (-3) = ...

Ghép cặp với a - 1 và 2 - b là ra :)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\frac{a}{5}=\frac{b}{4}=\frac{c}{5}=\frac{2a}{10}=\frac{3b}{12}=\frac{4c}{20}=\frac{2a-3b-4c}{10-12-20}=\frac{28}{-22}=\frac{-14}{11}\)

\(\Rightarrow\hept{\begin{cases}a=\frac{-14}{11}.5=\frac{-70}{11}\\b=\frac{-14}{11}.4=\frac{-56}{11}\\c=\frac{-14}{11}.5=\frac{-70}{11}\end{cases}}\)