a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)

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

\(=\left(1-sin^2a\right)-sin^2a=1\)

b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)

\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)

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

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

c/ \(C=sin^6x+cos^6x+3sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3sin^2x.cos^2x\)

\(=sin^4x-sin^2x.cos^2x+cos^4x+3sin^2x.cos^2x\)

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

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

sin a=3/5

=>cos a=4/5

tan a=3/5:4/5=3/4; cot a=1:3/4=4/3

M=(4/3+3/4):(4/3-3/4)=25/7

\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)

\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)

\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)

1) \(cot\alpha=\sqrt[]{5}\Rightarrow tan\alpha=\dfrac{1}{\sqrt[]{5}}\)

\(C=sin^2\alpha-sin\alpha.cos\alpha+cos^2\alpha\)

\(\Leftrightarrow C=\dfrac{1}{cos^2\alpha}\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+tan^2\alpha\right)\left(tan^2\alpha-tan\alpha+1\right)\)

\(\Leftrightarrow C=\left(1+\dfrac{1}{5}\right)\left(\dfrac{1}{5}-\dfrac{1}{\sqrt[]{5}}+1\right)\)

\(\Leftrightarrow C=\dfrac{6}{5}\left(\dfrac{6}{5}-\dfrac{\sqrt[]{5}}{5}\right)=\dfrac{6}{25}\left(6-\sqrt[]{5}\right)\)

1: \(cota=\sqrt{5}\)

=>\(cosa=\sqrt{5}\cdot sina\)

\(1+cot^2a=\dfrac{1}{sin^2a}\)

=>\(\dfrac{1}{sin^2a}=1+5=6\)

=>\(sin^2a=\dfrac{1}{6}\)

\(C=sin^2a-sina\cdot\sqrt{5}\cdot sina+\left(\sqrt{5}\cdot sina\right)^2\)

\(=sin^2a\left(1-\sqrt{5}+5\right)=\dfrac{1}{6}\cdot\left(6-\sqrt{5}\right)\)

2: tan a=3

=>sin a=3*cosa 

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

=>\(\dfrac{1}{cos^2a}=1+9=10\)
=>\(cos^2a=\dfrac{1}{10}\)

\(B=\dfrac{3\cdot cosa-cosa}{27\cdot cos^3a+3\cdot cos^3a+2\cdot3\cdot cosa}\)

\(=\dfrac{2\cdot cosa}{30cos^3a+6cosa}=\dfrac{2}{30cos^2a+6}\)

\(=\dfrac{2}{3+6}=\dfrac{2}{9}\)

mik ko bít

I don't now

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a)     \({\cos ^2}\alpha  + {\sin ^2}\alpha  = 1\)

b)     \(\tan \alpha .\cot \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }}.\frac{{\cos \alpha }}{{\sin \alpha }} = 1\)

c)     \(\frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = {\tan ^2}\alpha  + 1\)

d)     \(\frac{1}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = 1 + {\cot ^2}\alpha \)