Prove that sec theta+tan theta=root(1+sin theta/1-sin theta)

To prove that \( \sec \theta + \tan \theta = \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}} \), we can start by expressing both sides in terms of sine and cosine. This will allow us to manipulate the equation more easily.

Understanding the Left Side

First, let’s rewrite \( \sec \theta \) and \( \tan \theta \) using sine and cosine:

• \( \sec \theta = \frac{1}{\cos \theta} \)
• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)

So, the left side becomes:

\( \sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \frac{1 + \sin \theta}{\cos \theta} \)

Examining the Right Side

Now let’s simplify the right side. We need to express \( \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}} \) in a form that we can compare to the left side. To do this, multiply the numerator and denominator by \( \frac{1 + \sin \theta}{1 + \sin \theta} \):

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

Using the Pythagorean identity \( 1 - \sin^2 \theta = \cos^2 \theta \), we can simplify further:

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

Bringing It All Together

Now we can see that both sides are identical:

Left side: \( \sec \theta + \tan \theta = \frac{1 + \sin \theta}{\cos \theta} \)

Right side: \( \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}} = \frac{1 + \sin \theta}{\cos \theta}

Thus, we have shown that \( \sec \theta + \tan \theta = \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}} \) is indeed true, confirming the equality through algebraic manipulation and the use of trigonometric identities.