Question

(cosa/1-sina+cosa/1+sina)*1/1+tg²a

Answer 1

Объяснение:

$(\frac{cos\alpha }{1-sin\alpha }+\frac{cos\alpha }{1+sin\alpha })*\frac{1}{1+tg^2\alpha } .$

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$3)\ \frac{2*sin\alpha }{cos\alpha } *cos^2\alpha =2*sin\alpha *cos\alpha =sin2\alpha .$

Answer 2

Interesting expression! It seems like a combination of trigonometric functions. If I understand correctly, you have $\frac{\cos a}{1 - \sin a} + \frac{\cos a}{1 + \sin a} \times \frac{1}{1 + \tan^2 a}$.

Let me simplify that for you.

Starting with the expression $\frac{\cos a}{1 - \sin a} + \frac{\cos a}{1 + \sin a} \times \frac{1}{1 + \tan^2 a}$:

First, let's find a common denominator for the fractions in the expression.

The common denominator for $(1 - \sin a)$ and $(1 + \sin a)$ is $(1 - \sin^2 a)$, which is equal to $\cos^2 a$.

Now, rewrite the expression with the common denominator:

$\frac{\cos a \cdot \cos^2 a}{\cos^2 a - \cos a \cdot \sin a} + \frac{\cos a \cdot \cos^2 a}{\cos^2 a + \cos a \cdot \sin a} \times \frac{1}{\cos^2 a}$

Combine the fractions:

$\frac{\cos a \cdot \cos^2 a + \cos a \cdot \cos^2 a}{\cos^2 a - \cos a \cdot \sin a} \times \frac{1}{\cos^2 a}$

Simplify the numerator:

$\frac{2 \cdot \cos a \cdot \cos^2 a}{\cos^2 a - \cos a \cdot \sin a} \times \frac{1}{\cos^2 a}$

Now, simplify further:

$\frac{2 \cdot \cos a}{1 - \frac{\sin a}{\cos a}}$

Using the identity $\tan a = \frac{\sin a}{\cos a}$:

$\frac{2 \cdot \cos a}{1 - \tan a}$

So, the simplified expression is $\frac{2 \cdot \cos a}{1 - \tan a}$.