Question

3+3tg(в квадрате)x - cos(в квадрате)x, если sinx=корень из 3/2

Answer 1

I hope this helps you

Answer 2

Решение во вложении---------------------

Answer 3

To find the value of the expression $3 + 3\tan^2(x) - \cos^2(x)$ when $\sin(x) = \sqrt{3}/2$, we can first determine the value of $\cos(x)$ using the given information.

We know that $\sin^2(x) + \cos^2(x) = 1$. Substituting the value of $\sin(x)$, we have:

$\left(\sqrt{3}/2\right)^2 + \cos^2(x) = 1$

$\frac{3}{4} + \cos^2(x) = 1$

$\cos^2(x) = 1 - \frac{3}{4}$

$\cos^2(x) = \frac{1}{4}$

Taking the square root of both sides, we get:

$\cos(x) = \pm \frac{1}{2}$

Since $\sin(x) = \sqrt{3}/2 > 0$ in the given range, we can determine that $\cos(x) = \frac{1}{2}$.

Now, substituting the value of $\cos(x)$ into the original expression:

$3 + 3\tan^2(x) - \cos^2(x)$

$3 + 3\tan^2(x) - \frac{1}{4}$

We can use the identity $\tan^2(x) = \sec^2(x) - 1$ to rewrite the expression:

$3 + 3(\sec^2(x) - 1) - \frac{1}{4}$

$3 + 3\sec^2(x) - 3 - \frac{1}{4}$

Simplifying further:

$3\sec^2(x) - \frac{1}{4}$

To find the value of $\sec(x)$, we can use the fact that $\sec^2(x) = 1 + \tan^2(x)$:

$\sec^2(x) = 1 + \left(\frac{1}{\cos^2(x)}\right)$

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

$\sec^2(x) = 1 + 4$

$\sec^2(x) = 5$

Taking the square root of both sides:

$\sec(x) = \sqrt{5}$

Now substituting this value back into the expression:

$3\sec^2(x) - \frac{1}{4}$

$3(\sqrt{5})^2 - \frac{1}{4}$

$3 \cdot 5 - \frac{1}{4}$