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Sửa đề: \(2\cdot sin\left(180-a\right)\cdot cota-cos\left(180-a\right)\cdot tana+cot\left(180-a\right)\)

\(=2\cdot sina\cdot cota+cosa\cdot tana+\dfrac{cos\left(180-a\right)}{sin\left(180-a\right)}\)

\(=2\cdot sina\cdot\dfrac{cosa}{sina}+cosa\cdot\dfrac{sina}{cosa}+\dfrac{-cosa}{sina}\)

\(=2cosa+sina-tana\)

21 tháng 10 2019

sin 100= sin 80

cos 16=có164

A=2sin 80+2cos16

B=2sin a*cot a-cos a*tan a*cot a

B=2sin a*cosa/sin a-cos a

B=2cos a-cos a=cos a

NV
4 tháng 3 2021

\(tana-cota=2\sqrt{3}\Rightarrow\left(tana-cota\right)^2=12\)

\(\Rightarrow\left(tana+cota\right)^2-4=12\Rightarrow\left(tana+cota\right)^2=16\)

\(\Rightarrow P=4\)

\(sinx+cosx=\dfrac{1}{5}\Rightarrow\left(sinx+cosx\right)^2=\dfrac{1}{25}\)

\(\Rightarrow1+2sinx.cosx=\dfrac{1}{25}\Rightarrow sinx.cosx=-\dfrac{12}{25}\)

\(P=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{sinx.cosx}=\dfrac{1}{-\dfrac{12}{25}}=-\dfrac{25}{12}\)

-4 ở đâu ra vậy ạ

 

22 tháng 10 2023

\(0< a< 90^0\)

=>\(sina>0\)

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

=>\(sin^2a=1-\dfrac{9}{16}=\dfrac{7}{16}\)

=>\(sina=\dfrac{\sqrt{7}}{4}\)

\(tana=\dfrac{sina}{cosa}=\dfrac{\sqrt{7}}{4}:\dfrac{3}{4}=\dfrac{\sqrt{7}}{3}\)

\(cota=\dfrac{1}{tana}=\dfrac{3}{\sqrt{7}}\)

\(A=\dfrac{tana+3cota}{tana+cota}=\dfrac{\dfrac{\sqrt{7}}{3}+\dfrac{9}{\sqrt{7}}}{\dfrac{3}{\sqrt{7}}+\dfrac{\sqrt{7}}{3}}\)

\(=\dfrac{34}{3\sqrt{7}}:\dfrac{16}{3\sqrt{7}}=\dfrac{17}{8}\)

NV
8 tháng 6 2020

\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)

\(=tan^2a+1=\frac{1}{cos^2a}\)

\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)

\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)

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

6 tháng 11 2019

a, \(\left(1-sin^2x\right)cot^2x+1-cot^2x\)

\(=cot^2x-sin^2x.cot^2x+1-cot^2x\)

\(=1-sin^2x.\frac{\text{cos}^2x}{sin^2x}=1-\text{cos}^2x=sin^2x\)

b,\(\left(tanx+cotx\right)^2-\left(tanx-cotx\right)2\)

\(=tan^2x2.tanx.cotx+cot^2x-tan^2x+2tanx.cotx-cot^2x\)

\(=4tanxcotx=4\)

c,\(\left(xsina-y\text{cos}a\right)^2+\left(x\text{cos}a+ysina\right)^2\)

\(=x^2sin^2a=2xysina\text{cos}a+y^2\text{cos}^2a+2xysina\text{cos}a+y^2sin^2a\)

\(=x^2\left(sin^2a+\text{cos}^2a\right)+y^2\left(sin^2a+\text{cos}^2a\right)\)

\(=x^2+y^2\)

18 tháng 7 2022

a) Ta có A=\dfrac{\tan \alpha+3 \dfrac{1}{\tan \alpha}}{\tan \alpha+\dfrac{1}{\tan \alpha}}=\dfrac{\tan ^{2} \alpha+3}{\tan ^{2} \alpha+1}=\dfrac{\dfrac{1}{\cos ^{2} \alpha}+2}{\dfrac{1}{\cos ^{2} \alpha}}=1+2 \cos ^{2} \alpha Suy ra A=1+2 \cdot \dfrac{9}{16}=\dfrac{17}{8}.

b) B=\dfrac{\dfrac{\sin \alpha}{\cos ^{3} \alpha}-\dfrac{\cos \alpha}{\cos ^{3} \alpha}}{\dfrac{\sin ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{3 \cos ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{2 \sin \alpha}{\cos ^{3} \alpha}}=\dfrac{\tan \alpha\left(\tan ^{2} \alpha+1\right)-\left(\tan ^{2} \alpha+1\right)}{\tan ^{3} \alpha+3+2 \tan \alpha\left(\tan ^{2} \alpha+1\right)}.

Suy ra B=\dfrac{\sqrt{2}(2+1)-(2+1)}{2 \sqrt{2}+3+2 \sqrt{2}(2+1)}=\dfrac{3(\sqrt{2}-1)}{3+8 \sqrt{2}}.

NV
24 tháng 4 2020

\(\frac{cos^2x-sin^2x}{cot^2x-tan^2x}-cos^2x=\frac{cos^2x-sin^2x}{\frac{cos^2x}{sin^2x}-\frac{sin^2x}{cos^2x}}-cos^2x\)

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

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

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