document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1165712>Question 1165712</A>: <I><SPAN STYLE=\"word-break: break-all;\"> For the function f(x) = tan x, show that f(x+y)<BR>\n" );
document.write( "- f(x)= sec^2 x tan y/1-tan x tan y </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852939 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=852939\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The equation you provided, $f(x+y) - f(x) = \frac{\sec^2 x \tan y}{1 - \tan x \tan y}$, is incorrect or requires a different approach to prove. The correct simplification for the left-hand side is generally simpler.\r<BR>\n" );
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document.write( "Here is the correct derivation for $f(x+y) - f(x)$ and an explanation of the identity you provided.\r<BR>\n" );
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document.write( "## 1. Correct Derivation for $f(x+y) - f(x)$\r<BR>\n" );
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document.write( "Given $f(x) = \tan x$, the expression $f(x+y) - f(x)$ is:\r<BR>\n" );
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document.write( "$$f(x+y) - f(x) = \tan(x+y) - \tan x$$\r<BR>\n" );
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document.write( "Using the tangent addition formula, $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$:\r<BR>\n" );
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document.write( "$$\tan(x+y) - \tan x = \frac{\tan x + \tan y}{1 - \tan x \tan y} - \tan x$$\r<BR>\n" );
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document.write( "To combine the terms, find a common denominator:\r<BR>\n" );
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document.write( "$$\frac{\tan x + \tan y}{1 - \tan x \tan y} - \frac{\tan x (1 - \tan x \tan y)}{1 - \tan x \tan y}$$\r<BR>\n" );
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document.write( "Combine the numerators:\r<BR>\n" );
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document.write( "$$\frac{(\tan x + \tan y) - (\tan x - \tan^2 x \tan y)}{1 - \tan x \tan y}$$\r<BR>\n" );
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document.write( "Distribute the negative sign in the numerator:\r<BR>\n" );
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document.write( "$$\frac{\tan x + \tan y - \tan x + \tan^2 x \tan y}{1 - \tan x \tan y}$$\r<BR>\n" );
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document.write( "The $\tan x$ terms cancel out:\r<BR>\n" );
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document.write( "$$\frac{\tan y + \tan^2 x \tan y}{1 - \tan x \tan y}$$\r<BR>\n" );
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document.write( "Factor out $\tan y$ from the numerator:\r<BR>\n" );
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document.write( "$$\frac{\tan y (1 + \tan^2 x)}{1 - \tan x \tan y}$$\r<BR>\n" );
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document.write( "Finally, use the Pythagorean identity $\mathbf{1 + \tan^2 x = \sec^2 x}$:\r<BR>\n" );
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document.write( "$$f(x+y) - f(x) = \frac{\sec^2 x \tan y}{1 - \tan x \tan y}$$\r<BR>\n" );
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document.write( "## 2. Conclusion\r<BR>\n" );
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document.write( "The identity you asked to show, $\mathbf{f(x+y) - f(x) = \frac{\sec^2 x \tan y}{1 - \tan x \tan y}}$, is **correct**.\r<BR>\n" );
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document.write( "Your expression:<BR>\n" );
document.write( "$$\mathbf{LHS: f(x+y) - f(x)}$$\r<BR>\n" );
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document.write( "Your proposed result:<BR>\n" );
document.write( "$$\mathbf{RHS: \frac{\sec^2 x \tan y}{1 - \tan x \tan y}}$$\r<BR>\n" );
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document.write( "Since the derivation of $f(x+y) - f(x)$ leads directly to $\frac{\sec^2 x \tan y}{1 - \tan x \tan y}$, the identity is proven. </SPAN>\n" );
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