Ta có: \(\tan x\cdot\cot x=1\Leftrightarrow\tan x\cdot\frac{8}{15}=1\Leftrightarrow\tan x=1:\frac{8}{15}=\frac{15}{8}\)

Lại có: \(\tan x=\frac{\sin x}{\cos x}=\frac{15}{8}\Leftrightarrow\sin x=\frac{15}{8}\cos x\)

Mặt khác:

\(\sin^2x+\cos^2x=1\\ \Leftrightarrow\left(\frac{15}{8}\cos x\right)^2+\cos^2x=1\\ \Leftrightarrow\frac{225}{64}\cos^2x+\cos^2x=1\\ \Leftrightarrow\frac{289}{64}\cos^2x=1\\ \Leftrightarrow\cos^2x=\frac{64}{289}\\ \Leftrightarrow\cos x=\frac{8}{17}\)

Suy ra: \(\sin x=\frac{15}{8}\cdot\cos x=\frac{15}{8}\cdot\frac{8}{17}=\frac{15}{17}\)

a) Ta có: \(\sin^2a^o=\cos^2\left(90^o-a^o\right)\)

Biểu thức trên

\(=\left(\sin^21^o+\sin^o89\right)+\left(\sin^22^o+\sin^288^o\right)+...+\left(\sin^244^o+\sin^246^o\right)+\sin^245^o\)

\(=\left(\sin^21^o+\cos^21^o\right)+\left(\sin^22^o+\cos^22^o\right)+...+\left(\sin^244^o+\cos^246^o\right)+\sin^245^o\)

\(=1+1+..+1+\sin^245^o=44+\frac{1}{2}=\frac{89}{2}\)

b) 

Ta có: \(\sin^2x+\cos^2x=1\)

\(0^o< x< 90^o\)

=> \(0< \sin x;\cos x< 1\)

Ta có:  \(\frac{\sin^2x+\cos^2x}{\text{​​}\text{​​}\sin x.\cos x}=\frac{1}{\frac{12}{25}}=\frac{25}{12}\Leftrightarrow\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{25}{12}\)

\(\Leftrightarrow\tan x+\frac{1}{\tan x}=\frac{25}{12}\Leftrightarrow\tan^2x-\frac{25}{12}\tan x+1=0\)

Đặt t =tan x => có phương trình bậc 2 ẩn t => Giải đen ta => ra đc t => ra đc tan t

\(\Leftrightarrow\orbr{\begin{cases}\tan x=\frac{3}{4}\\\tan x=\frac{4}{3}\end{cases}}\)

\(x=45^0\)

a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow x=30^o\)

b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=30^o\)

c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)

d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)

Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(

e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)

f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)

a) \(\frac{1}{cos^2x}=1+tan^2x=1+\frac{9}{16}=\frac{25}{16}\)

\(\Leftrightarrow cos^2x=\frac{16}{25}\Leftrightarrow\orbr{\begin{cases}cosx=\frac{4}{5}\\cosx=\frac{-4}{5}\end{cases}}\)

- \(cosx=\frac{4}{5}\): 

\(sinx=cosxtanx=\frac{4}{5}.\frac{3}{4}=\frac{3}{5}\)

\(cotx=\frac{1}{tanx}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\).

- \(cosx=\frac{-4}{5}\): 

\(sinx=cosxtanx=\frac{-4}{5}.\frac{3}{4}=\frac{-3}{5}\)

\(cotx=\frac{1}{tanx}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\).

b)  \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{49}{625}=\frac{576}{625}\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{24}{25}\\cosx=-\frac{24}{25}\end{cases}}\)

- \(cosx=\frac{24}{25}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{7}{25}}{\frac{24}{25}}=\frac{7}{24}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\frac{7}{24}}=\frac{24}{7}\)

- \(cosx=\frac{-24}{25}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{7}{25}}{\frac{-24}{25}}=-\frac{7}{24}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\frac{7}{24}}=\frac{-24}{7}\)