Question

Let 0π4.Then (sec2x−tan2x) equal
(1)tan(π−π4)
(2)tan(π4−x)
(3)tan(x+π4)
(4)tan2(x+π4)

Answer 1

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]