How do you simplify the expression $sec (arctan (\frac{2 x}{5}))$ $sec(arctan ((2x)/5))$ ?

Question

1904 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-22T12:25:18

$\sqrt{1 + \frac{4}{15}} x^{2}$ $sqrt(1+4/15)x^2$

Let $a = a r c tan ((\frac{2}{5}) x) \in Q 1 or Q 4$ $a = arc tan ((2/5)x) in Q1 or Q4$ , according as $x > or < 0$ $x > or < 0$ .

Then, $tan a = \frac{2}{5} x, and sec > 0$ $tan a = 2/5x, and sec > 0$ , in both Q1 and Q4.

Now, the given expression is

$sec a = \sqrt{1 + {tan}^{2} a} = \sqrt{1 + \frac{4}{15}} x^{2}$ $sec a = sqrt(1+tan^2a)=sqrt(1+4/15)x^2$ .

If a is assumed to be the general value, it might $\in Q 3$ $in Q3$ , for negative

x. In this case, sec a becomes negative and the answer becomes

$\pm \sqrt{1 + \frac{4}{25} x^{2}},$ $+-sqrt(1+4/25x^2),$ ..

How do you simplify the expression sec(arctan(2x5))sec(arctan ((2x)/5))?