\(tan^2x+cot^2x=2=2.tanx.cotx\)

\(\Leftrightarrow tan^2x+cot^2x-2tanx.cotx=0\)

\(\Leftrightarrow\left(tanx-cotx\right)^2=0\Leftrightarrow tanx=cotx=\dfrac{1}{tanx}\)

\(\Leftrightarrow tanx=\pm1\)

\(P=\dfrac{1}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{1+sinx-cos^2x}{cosx\left(1+sinx\right)}=\dfrac{sin^2x+sinx}{cosx\left(1+sinx\right)}\)

\(=\dfrac{sinx\left(1+sinx\right)}{cosx\left(1+sinx\right)}=tanx=\pm1\)

a: Sửa đề: sin x=4/5

cosx=-3/5; tan x=-4/3; cot x=-3/4

b: 270 độ<x<360 độ

=>cosx>0

=>cosx=1/2

tan x=căn 3; cot x=1/căn 3

\(90^0< a< 180^0\)

=>\(cosa< 0\)

\(sin^2a+cos^2a=1\)

=>\(cos^2a+\dfrac{9}{25}=1\)

=>\(cos^2a=1-\dfrac{9}{25}=\dfrac{16}{25}\)

mà cosa<0

nên \(cosa=-\dfrac{4}{5}\)

\(tana=\dfrac{sina}{cosa}=\dfrac{3}{5}:\dfrac{-4}{5}=-\dfrac{3}{4}\)

\(A=2\cdot cos^2a-5\cdot tan^2a\)

\(=2\cdot\left(-\dfrac{4}{5}\right)^2-5\cdot\left(-\dfrac{3}{4}\right)^2\)

\(=2\cdot\dfrac{16}{25}-5\cdot\dfrac{9}{16}\)

\(=\dfrac{32}{25}-\dfrac{45}{16}=\dfrac{-613}{400}\)

\(\dfrac{\pi}{2}< x< \pi\Rightarrow sinx>0\)

\(\Rightarrow sinx=\sqrt{1-cos^2x}=\sqrt{1-\left(-\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)

\(sin\left(x+\dfrac{\pi}{3}\right)=sinx.cos\left(\dfrac{\pi}{3}\right)+cosx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{4}{5}.\dfrac{1}{2}+\left(-\dfrac{3}{5}\right).\dfrac{\sqrt{3}}{2}=\dfrac{4-3\sqrt{3}}{10}\)

\(cos\left(x+\dfrac{\pi}{4}\right)=cosx.cos\left(\dfrac{\pi}{4}\right)-sinx.sin\left(\dfrac{\pi}{4}\right)=-\dfrac{3}{5}.\dfrac{\sqrt{2}}{2}-\dfrac{4}{5}.\dfrac{\sqrt{2}}{2}=-\dfrac{7\sqrt{2}}{10}\)

\(sin(\dfrac{\pi}{2}-x)cot(\pi+x)=cosxcotx=\dfrac{cosx}{tanx}\\ =\dfrac{\dfrac{1}{\sqrt5}}{-2}=\dfrac{-\sqrt5}{10}\)