Question

Sin a/1+cos a - sin a/1-cos a = - 2 ctg a​

Answer 1

Ответ:

$\dfrac{sina}{1+cosa}-\dfrac{sina}{1-cosa}=\dfrac{sina(1-cosa)-sina(1+cosa)}{(1+cosa)(1-cosa)}=\\\\\\=\dfrac{sina-sina\cdot cosa-sina-sina\cdot cosa}{(1+cosa)(1-cosa)}=\dfrac{-2sina\cdot cosa}{1-cos^2a}=\\\\\\=-\dfrac{2sina\cdot cosa}{sin^2a}=-\dfrac{2\cdot cosa}{sina}=-2\cdot ctga\\\\\\-2ctga=-2ctga$

Answer 2

Let's simplify the given expression step by step:

$\frac{\sin a}{1 + \cos a} - \frac{\sin a}{1 - \cos a}$

To combine these fractions, let's find a common denominator, which is the product of the denominators $(1 + \cos a)(1 - \cos a)$:

$\frac{\sin a(1 - \cos a)}{(1 + \cos a)(1 - \cos a)} - \frac{\sin a(1 + \cos a)}{(1 + \cos a)(1 - \cos a)}$

Now, combine the fractions:

$\frac{\sin a(1 - \cos a) - \sin a(1 + \cos a)}{(1 + \cos a)(1 - \cos a)}$

Distribute $\sin a$ in both terms of the numerator:

$\frac{\sin a - \sin a \cos a - \sin a - \sin a \cos a}{(1 + \cos a)(1 - \cos a)}$

Combine like terms in the numerator:

$\frac{-2 \sin a \cos a}{(1 + \cos a)(1 - \cos a)}$

Now, let's simplify the expression further. Recall that $\cos^2 a = 1 - \sin^2 a$, so $\cos a = \sqrt{1 - \sin^2 a}$. Substitute this into the expression:

$\frac{-2 \sin a \sqrt{1 - \sin^2 a}}{(1 + \sqrt{1 - \sin^2 a})(1 - \sqrt{1 - \sin^2 a})}$

Now, notice that the denominators have a difference of squares, $a^2 - b^2 = (a + b)(a - b)$. In this case, let $a = 1$ and $b = \sqrt{1 - \sin^2 a}$:

$\frac{-2 \sin a \sqrt{1 - \sin^2 a}}{(1 + \sqrt{1 - \sin^2 a})(1 - \sqrt{1 - \sin^2 a})}$

$= \frac{-2 \sin a \sqrt{1 - \sin^2 a}}{(1 + \sqrt{1 - \sin^2 a})(1 - \sqrt{1 - \sin^2 a})} \cdot \frac{1 + \sqrt{1 - \sin^2 a}}{1 + \sqrt{1 - \sin^2 a}}$