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Answer» Math]secx + TANX[/math]
[math]secx = \dfrac{1}{cosx}[/math]
And we know that
[math]cosx = \dfrac{1-TAN^2\frac{x}{2}}{1+tan^2\frac{x}{2}}[/math]
So,
[math]secx = \dfrac{1+tan^2\frac{x}{2}}{1-tan^2\frac{x}{2}}[/math]
ALSO
[math]tanx = \dfrac{2tan\frac{x}{2}}{1-tan^2\frac{x}{2}}[/math]
On adding the 2 equations we get
[math]secx + tanx = \dfrac{1+2tan\frac{x}{2}+tan^2\frac{x}{2}}{1-tan^2\frac{x}{2}}[/math]
LHS = [math]\dfrac{(1+tan\frac{x}{2})^2}{(1-tan\frac{x}{2})(1+tan{x}{2})}[/math]
Cancelling one of the terms of 1+tan(x/2) we get
LHS = [math]\dfrac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}[/math]
LHS = [math]tan (\frac{\pi}{4} + \frac{x}{2})[/math]
So we see that we get a TERM of [math]\frac{\pi}{4}[/math] and not [math]\frac{\pi}{8}[/math]
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