Question

[tg(180º+x)]: [ctg(90º-x) ] + tg²x ​

Answer 1

$\frac{tg(180^{0}+x) }{Ctg(90^{0}-x) }+tg^{2} x=\frac{tgx}{tgx} +tg^{2}x=1+tg^{2}x=\boxed{\frac{1}{Cos^{2}x }}$

Answer 2

To simplify the given expression, we'll use trigonometric identities. First, we'll express all trigonometric functions in terms of sine and cosine, and then apply the identities.

The basic trigonometric identities are:

1. $\tan(x) = \frac{\sin(x)}{\cos(x)}$
2. $\cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}$

We also know the following identities:

1. $\tan(x) = -\cot(180^\circ + x)$
2. $\cot(x) = -\tan(90^\circ - x)$

Let's use these to simplify the expression:

$\tan(180^\circ + x) = -\cot(x)$ $\cot(90^\circ - x) = -\tan(x)$

Thus, the expression becomes:

$\left[-\cot(x)\right] + \tan^2(x)$

We know that $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and $\cot(x) = \frac{\cos(x)}{\sin(x)}$, so we can substitute these back into the expression:

$\left[-\frac{\cos(x)}{\sin(x)}\right] + \frac{\sin^2(x)}{\cos^2(x)}$

Now, we'll find a common denominator:

$\frac{-\cos^2(x) + \sin^2(x)}{\sin(x)\cos^2(x)}$

By using the identity $\cos^2(x) + \sin^2(x) = 1$, we simplify further:

$\frac{-1}{\sin(x)\cos^2(x)}$

So, the simplified expression is $-\csc(x)\sec^2(x)$.