Bài 9.
(a) \(VT=1+tan^2\alpha=1+\dfrac{sin^2\alpha}{cos^2\alpha}=\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}\)
\(=\dfrac{1}{cos^2\alpha}=VP\left(đpcm\right)\)
(b) \(VT=1+cot^2\alpha=1+\dfrac{cos^2\alpha}{sin^2\alpha}=\dfrac{sin^2\alpha+cos^2\alpha}{sin^2\alpha}=\dfrac{1}{sin^2\alpha}=VP\left(đpcm\right)\)
(c) \(VT=\left(sin\alpha+cos\alpha\right)^2=sin^2\alpha+cos^2\alpha+2sin\alpha.cos\alpha=1+2sin\alpha.cos\alpha=VP\left(đpcm\right)\)
(d) \(VT=\left(tan\alpha+cot\alpha\right)^2=tan^2\alpha+cot^2\alpha+2tan\alpha.cot\alpha\)
\(=tan^2\alpha+cot^2\alpha+2\cdot\dfrac{sin\alpha}{cos\alpha}\cdot\dfrac{cos\alpha}{sin\alpha}\)
\(=tan^2\alpha+cot^2\alpha+2=VP\left(đpcm\right)\)
(e) \(VT=\dfrac{sin^3x+cos^2x}{sin^2x}=sinx+\dfrac{cos^2x}{sin^2x}=sinx+cot^2x=VP\left(đpcm\right)\)
(f) \(VT=\dfrac{1+cotx}{1-cotx}=\dfrac{1+\dfrac{cosx}{sinx}}{1-\dfrac{cosx}{sinx}}\)
\(=\dfrac{\dfrac{sinx+cosx}{sinx}}{\dfrac{sinx-cosx}{sinx}}=\dfrac{sinx+cosx}{sinx-cosx}\)
Lại có: \(VP=\dfrac{tanx+1}{tanx-1}=\dfrac{\dfrac{sinx}{cosx}+1}{\dfrac{sinx}{cosx}-1}\)
\(=\dfrac{\dfrac{sinx+cosx}{cosx}}{\dfrac{sinx-cosx}{cosx}}=\dfrac{sinx+cosx}{sinx-cosx}\)
\(\Rightarrow VT=VP=\dfrac{sinx+cosx}{sinx-cosx}\left(đpcm\right)\)
(g) \(VT=cos^4x-sin^4x\)
\(=-\left(sin^4x-cos^4x\right)\)
\(=-\left[\left(sin^4x+2sin^2x.cos^2x+cos^4x\right)-2cos^4x-2sin^2x.cos^2x\right]\)
\(=-\left[\left(sin^2x+cos^2x\right)^2-2cos^2x\left(sin^2x+cos^2x\right)\right]\)
\(=-\left(1-2cos^2x\right)=2cos^2x-1=VP\left(dpcm\right)\)
(h) \(VT=sin^4x+cos^4x\)
\(=\left(sin^4x+2sin^2x.cos^2x+cos^4x\right)-2sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)
\(=1-2sin^2x.cos^2x=VP\left(dpcm\right)\)
10:
a: 90<a<180
=>cos a<0
=>\(cosa=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)
\(tana=\dfrac{3}{5}:\dfrac{-4}{5}=\dfrac{-3}{4}\)
cot a=1/tan a=-4/3
b: 0<a<pi/2
=>sin a>0
=>\(sina=\sqrt{1-\left(\dfrac{2}{3}\right)^2}=\dfrac{\sqrt{5}}{3}\)
\(tana=\dfrac{\sqrt{5}}{3}:\dfrac{2}{3}=\dfrac{\sqrt{5}}{2}\)
\(cota=1:\dfrac{\sqrt{5}}{2}=\dfrac{2}{\sqrt{5}}\)
c: pi/2<a<pi
=>sina>0 và cosa<0
\(1+tan^2a=\dfrac{1}{cos^2a}=1+5=6\)
=>\(cos^2a=\dfrac{1}{6}\)
mà cos a<0
nên \(cosa=-\dfrac{1}{\sqrt{6}}\)
=>\(sina=\sqrt{1-\dfrac{1}{6}}=\sqrt{\dfrac{5}{6}}=\dfrac{\sqrt{30}}{6}\)
\(cota=\dfrac{1}{tana}=\dfrac{-1}{\sqrt{5}}\)
d: pi/2<a<pi
=>cosa<0 và sin a>0
\(1+cot^2a=\dfrac{1}{sin^2a}\)
=>\(\dfrac{1}{sin^2a}=1+\dfrac{1}{2}=\dfrac{3}{2}\)
=>sin^2a=2/3
=>\(sina=\sqrt{\dfrac{2}{3}}=\dfrac{\sqrt{6}}{3}\)
\(cosa=-\sqrt{1-\dfrac{2}{3}}=-\dfrac{1}{\sqrt{3}}\)
\(tana=\dfrac{1}{cota}=-\sqrt{2}\)
