Question

[Le Minh Dang]

Answer 1

[Tô Mì]

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) \(VT=sin^4x+cos^4x\)

\(=\left(sin^4x+2sin^2x.cos^2x+cos^4x\right)-2sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)

\(=1-2sin^2x.cos^2x=VP\left(dpcm\right)\)

Answer 2

[Nguyễn Lê Phước Thịnh]

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

=>\(sina=\sqrt{\dfrac{2}{3}}=\dfrac{\sqrt{6}}{3}\)

\(cosa=-\sqrt{1-\dfrac{2}{3}}=-\dfrac{1}{\sqrt{3}}\)

\(tana=\dfrac{1}{cota}=-\sqrt{2}\)