Question

Sin^2a/sina-cosa + sina+cosa/1-tg^2a=sina+cosa

Answer 1

Решение прикреплено. Удачи

Answer 2

To prove the trigonometric identity:

$\frac{\sin^2a}{\sin a - \cos a} + \frac{\sin a + \cos a}{1 - \tan^2a} = \sin a + \cos a$

We'll start with the left-hand side (LHS) and manipulate it step by step to reach the right-hand side (RHS).

Step 1: Factor out common terms. $\frac{\sin^2a}{\sin a - \cos a} + \frac{\sin a + \cos a}{1 - \tan^2a}$

Step 2: Write $\tan^2a$ in terms of $\sin a$ and $\cos a$. $\tan^2a = \frac{\sin^2a}{\cos^2a}$

Step 3: Rewrite the denominator $\cos^2a$ as $(1 - \sin^2a)$. $\tan^2a = \frac{\sin^2a}{1 - \sin^2a}$

Step 4: Substitute the rewritten $\tan^2a$ into the second fraction. $\frac{\sin^2a}{\sin a - \cos a} + \frac{\sin a + \cos a}{1 - \frac{\sin^2a}{1 - \sin^2a}}$

Step 5: Simplify the second fraction's denominator. $\frac{\sin^2a}{\sin a - \cos a} + \frac{\sin a + \cos a}{1 - \frac{\sin^2a}{\cos^2a}}$

Step 6: Rewrite $\cos^2a$ as $(1 - \sin^2a)$. $\frac{\sin^2a}{\sin a - \cos a} + \frac{\sin a + \cos a}{1 - \frac{\sin^2a}{1 - \sin^2a}}$

Step 7: Find a common denominator for the fractions in the expression. The common denominator is $(\sin a - \cos a)(1 - \sin^2a)$.

Step 8: Write each fraction with the common denominator. $\frac{\sin^2a(1 - \sin^2a)}{(\sin a - \cos a)(1 - \sin^2a)} + \frac{(\sin a + \cos a)(\sin a - \cos a)}{(\sin a - \cos a)(1 - \sin^2a)}$

Step 9: Combine the fractions. $\frac{\sin^2a(1 - \sin^2a) + (\sin a + \cos a)(\sin a - \cos a)}{(\sin a - \cos a)(1 - \sin^2a)}$

Step 10: Simplify the numerator. $\frac{\sin^2a - \sin^4a + \sin^2a - \cos^2a}{(\sin a - \cos a)(1 - \sin^2a)}$