Giải các phương trình sau: 1
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Note that $ an(x)=sin(x)/cos(x)=1/cot(x)$ and$$fraccos(x)1 - an(x)+ fracsin(x)1 -cot(x)=fraccos^2(x)cos(x) - sin(x)+ fracsin^2(x)sin(x) -cos(x)=fracsin^2(x)-cos^2(x)sin(x) -cos(x).$$ Can you take it from here?
You said you spent hours on this problem, it is possible you get confused about all these trigonometric functions. One suggestion could be to replace $cos(x)$ by $c$ và $sin(x)$ by $s$.
If this $quaddfracc1-frac sc+dfracs1-frac cs=dfracc^2c-s+dfracs^2s-c=dfracc^2-s^2c-s=dfrac(c-s)(c+s)c-s=c+s$
is making more sense to you, even though it"s just Robert.Z"s answer written with simpler symbols, then go for this sort of reduction when you work with trigonometric formulas.
Myself for instance I always write $c^2+s^2=1$ instead of $cos^2(x)+sin^2(x)=1$, I find it easier to lớn remember.
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$1- cot(x)=frac sin(x) -cos(x)sin(x)$
$1- tan(x)=frac cos(x) - sin(x)cos(x)$
Then we have
$fraccos^2(x) sin(x) -cos(x)+frac sin^2(x) sin(x) -cos(x)$
$ =frac-cos^2(x) +sin^2(x) sin(x) -cos(x)$=$frac(sin(x)-cos(x))(sin(x)+cos(x)) sin(x) -cos(x)$=$sin(x) +cos(x)$
note that $$fraccos(x)1- an(x)+fracsin(x)1-cot(x)-sin(x)-cos(x)=-cos left( x ight) an left( x ight) cot left( x ight) - an left( x ight) sin left( x ight) cot left( x ight) +cos left( x ight) an left( x ight) +sin left( x ight) cot left( x ight) $$ and cảnh báo that $$ an(x)cot(x)=1,cos(x) an(x)=sin(x),sin(x)cot(x)=cos(x)$$
Trigonometry Question: How to prove $arctan(cot^2(x)) + ccevents.vnrmarccot( an^2 (x)) = fracpi2$?
Prove $(1 + sin x + cos x -cos^2x sin x + sin^2x cos x )/ (cos x sin x) = (1+ sin^3x + cos^3x ) / (cos x sin x)$
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