Question

If tanØ =40/4 then fjnd secØ and cosØ​

Answer 1

Correct Question:-

• If $\sf \tan(x) = \frac{40}{9}$, then find $\sf \sec(x)$ and $\sf \cos(x)$

Solution:-

Given,

$\sf \tan(x) = \frac{40}{9}$

We know that,

$\sf { \sec }^{2} (x) = 1 + { \tan}^{2} (x)$

From this relation,, we get,

$\sf \implies \sec(x) = \sqrt{1 + { \tan }^{2}(x) }$

Now, putting the value, we get,

$\sf \implies \sec(x) = \sqrt{1 + \frac{1600}{81} }$

$\sf \implies \sec(x) = \sqrt{ \frac{1600 + 81}{81} }$

$\sf \implies \sec(x) = \sqrt{ \frac{1681}{81} }$

$\sf \implies \sec(x) = \frac{41}{9}$

Again, we know that,

$\sf \cos(x) = \frac{1 }{ \sec(x) }$

So,

$\sf \implies \cos(x) = \frac{1 }{{}^{41} / {}_{9} }$

$\sf \implies \cos(x) = \frac{9}{41}$

Answer:-

1. $\sf \sec(x) = \frac{41}{9}$
2. $\sf \cos(x) = \frac{9}{41}$