Question 1130046
<pre>The inverse cosine is taken in QI and QII between 0 and <font face="symbol">p</font>.

{{{drawing(70,40,0,2,-1,1.7,locate(0,.9,"cos"^(-1)),locate(1.4,.4,(0)) )}}}{{{""=""}}}{{{pi/2}}}

First find:

{{{drawing(80,40,0,2,-1,1.7,locate(0,.9,"cos"^(-1)),locate(1.2,1.3,(4pi/3)) )}}}{{{""=""}}}{{{-1/2}}}

Then

{{{"cos"^(-1)}}}{{{(cos(4pi/3)^"")}}}{{{""=""}}}{{{"cos"^(-1)}}}{{{-1/2}}}{{{""=""}}}{{{2pi/3}}}

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{{{expr(cos(x)/sin(x))*(tan(x)^"" + cos(x))}}} 

{{{expr(cos(x)/sin(x))*(sin(x)/cos(x) + cos(x)^""^"")}}}

Distribute

{{{expr(cos(x)/sin(x))*expr(sin(x)/cos(x)) + expr(cos(x)/sin(x))cos(x)}}}

{{{expr(cross(cos(x))/cross(sin(x)))*expr(cross(sin(x))/cross(cos(x))) + cos^2(x)/sin^""(x)}}}

{{{1+ cos^2(x)/sin^""(x)}}}

{{{1+ (1-sin^2(x))/sin^""(x)}}}

{{{1+ 1/sin(x)-sin^2(x)/sin^""(x)}}}

{{{1+ csc(x)-sin^cross(2)(x)/cross(sin^""(x))}}}

{{{1+csc(x)-sin(x)}}}

-----------------------------------

{{{(sin^2(x)^""^"" - 1)*(tan^2(x)^""^"" - 1)}}}

{{{(-1+sin^2(x)^""^"")*(tan^2(x)^""^"" - 1)}}}

{{{-1(1-sin^2(x)^""^"")*(tan^2(x)^""^"" - 1)}}}

{{{-1(cos^2(x)^""^"")*(tan^2(x)^""^"" - 1)}}}

{{{-cos^2(x)(tan^2(x)^""^"" - 1)}}}

{{{-cos^2(x)(sin^2(x)/cos^2(x) - 1)}}}

Distribute:

{{{-cos^2(x)expr(sin^2(x)/cos^2(x)) + cos^2(x)}}}

{{{-cross(cos^2(x))expr(sin^2(x)/cross(cos^2(x))) + cos^2(x)}}}

{{{-sin^2(x)+cos^2(x)}}}

{{{cos^2(x)-sin^2(x)}}}

{{{cos(2x)}}}

Edwin<pre>