$$ \frac{1-cos2a}{sin2a}=tana $$

Question

UpStudy AI · Accepted Answer

Yes, the equation you provided is a valid trigonometric identity. Let's verify and simplify the expression step by step to show that:

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

### Step 1: Use Double Angle Identities

Recall the double angle identities for sine and cosine:

$$
\cos 2a = 1 - 2\sin^2 a \quad \text{and} \quad \sin 2a = 2\sin a \cos a
$$

### Step 2: Substitute the Double Angle Identity for Cosine

Substitute $$ \cos 2a = 1 - 2\sin^2 a $$ into the numerator:

$$
1 - \cos 2a = 1 - (1 - 2\sin^2 a) = 2\sin^2 a
$$

So the expression becomes:

$$
\frac{2\sin^2 a}{\sin 2a}
$$

### Step 3: Substitute the Double Angle Identity for Sine

Substitute $$ \sin 2a = 2\sin a \cos a $$ into the denominator:

$$
\frac{2\sin^2 a}{2\sin a \cos a} = \frac{\sin^2 a}{\sin a \cos a}
$$

### Step 4: Simplify the Expression

Cancel out a $$ \sin a $$ from the numerator and denominator:

$$
\frac{\sin a}{\cos a} = \tan a
$$

### Conclusion

After simplifying, the original expression reduces to $$ \tan a $$. Therefore, the identity holds true:

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

\( \frac{1-cos2a}{sin2a}=tana \)