What is the answer of tan square tita- sin square tita=tan square tita× sin square tita

To solve the equation \( \tan^2 \theta - \sin^2 \theta = \tan^2 \theta \times \sin^2 \theta \), we can start by expressing everything in terms of sine and cosine. This will help us simplify the equation and find a solution.

Breaking Down the Equation

First, let's recall the definitions of tangent and sine:

• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
• \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \)

Now, we can substitute \( \tan^2 \theta \) into the equation:

Substituting and Rearranging

The equation becomes:

\( \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \times \sin^2 \theta \)

Next, let's factor out \( \sin^2 \theta \) from the left side:

\( \sin^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right) = \frac{\sin^4 \theta}{\cos^2 \theta} \)

Further Simplifying

We can simplify \( \frac{1}{\cos^2 \theta} - 1 \) as follows:

\( \frac{1 - \cos^2 \theta}{\cos^2 \theta} = \frac{\sin^2 \theta}{\cos^2 \theta} \)

Now we can substitute this back into the equation:

\( \sin^2 \theta \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right) = \frac{\sin^4 \theta}{\cos^2 \theta} \)

Final Steps to the Solution

At this point, we see that both sides of the equation are indeed equal:

\( \frac{\sin^4 \theta}{\cos^2 \theta} = \frac{\sin^4 \theta}{\cos^2 \theta} \)

This confirms that the original equation holds true for any angle \( \theta \) where both sine and cosine are defined. Thus, the equation simplifies correctly without any restrictions imposed on \( \theta \). Therefore, the solution is valid for all angles where \( \tan^2 \theta \) and \( \sin^2 \theta \) are defined.

Conclusion

In summary, we found that the equation \( \tan^2 \theta - \sin^2 \theta = \tan^2 \theta \times \sin^2 \theta \) holds true universally, provided that both sine and cosine are defined. This means that any angle \( \theta \) in the domain of these functions will satisfy the equation.