document.write( "Question 756834: Can someone please explain this equation to me? and i also believe that the answer would be b.
\n" ); document.write( "a=sin^-1 -x then a=
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\n" ); document.write( "a. cos^-1 1/sqrt 1-x^2
\n" ); document.write( "b. tan^-1 -x/sqrt 1-x^2
\n" ); document.write( "c. pi/2-cos 1/-x
\n" ); document.write( "d. tan^-1 sqrt 1+x^2/-x
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Algebra.Com's Answer #460607 by lwsshak3(11628)\"\" \"About 
You can put this solution on YOUR website!
a=sin^-1 -x then a=\r
\n" ); document.write( "\n" ); document.write( "a. cos^-1 1/sqrt 1-x^2
\n" ); document.write( "b. tan^-1 -x/sqrt 1-x^2
\n" ); document.write( "c. pi/2-cos 1/-x
\n" ); document.write( "d. tan^-1 sqrt 1+x^2/-x
\n" ); document.write( "..
\n" ); document.write( "a=sin^-1 -x
\n" ); document.write( "This expression reads:
\n" ); document.write( "an angle (a) whose sin=-x
\n" ); document.write( "In other words,
\n" ); document.write( "sin(a)=-x
\n" ); document.write( "cos(a)=sqrt(1-sin^2(a)=sqrt(1-x^2)
\n" ); document.write( "tan(a)=sin(a)/cos(a)=-x/sqrt(1-x^2)
\n" ); document.write( "this reads:
\n" ); document.write( "angle(a) whose tangent=-x/sqrt(1-x^2) (ans b.)
\n" ); document.write( "Remember: the inverse is always an angle
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