Simplify the expression:

$$\sqrt{\frac{\sec heta + 1}{\sec heta - 1}} + \sqrt{\frac{\sec heta - 1}{\sec heta + 1}}$$

Question

Simplify the expression:

$$\sqrt{\frac{\sec 	heta + 1}{\sec 	heta - 1}} + \sqrt{\frac{\sec 	heta - 1}{\sec 	heta + 1}}$$

Accepted Answer

**Solution:**

1. Start with the given expression:
   $$E = \sqrt{\frac{\sec \theta + 1}{\sec \theta - 1}} + \sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}}$$

2. Recall that $\sec \theta = \frac{1}{\cos \theta}$. Substitute this:
   $$E = \sqrt{\frac{\frac{1}{\cos \theta} + 1}{\frac{1}{\cos \theta} - 1}} + \sqrt{\frac{\frac{1}{\cos \theta} - 1}{\frac{1}{\cos \theta} + 1}}$$

3. Simplify numerator and denominator inside the square roots by getting common denominators:
   - For the first term numerator:
     $$\frac{1}{\cos \theta} + 1 = \frac{1 + \cos \theta}{\cos \theta}$$
   - For the first term denominator:
     $$\frac{1}{\cos \theta} - 1 = \frac{1 - \cos \theta}{\cos \theta}$$

So the first term inside the root becomes:
   $$\frac{\frac{1 + \cos \theta}{\cos \theta}}{\frac{1 - \cos \theta}{\cos \theta}} = \frac{1 + \cos \theta}{1 - \cos \theta}$$

4. Similarly, the second term inside the root is:
   $$\frac{\frac{1}{\cos \theta} - 1}{\frac{1}{\cos \theta} + 1} = \frac{1 - \cos \theta}{1 + \cos \theta}$$

5. So the expression reduces to:
   $$E = \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} + \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}$$

6. Use the identity:
   $$1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$$
   $$1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$$

7. Substitute these:
   $$E = \sqrt{\frac{2 \cos^2 \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}}} + \sqrt{\frac{2 \sin^2 \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}} = \sqrt{\frac{\cos^2 \frac{\theta}{2}}{\sin^2 \frac{\theta}{2}}} + \sqrt{\frac{\sin^2 \frac{\theta}{2}}{\cos^2 \frac{\theta}{2}}}$$

8. Simplify the square roots:
   $$E = \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} + \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} = \cot \frac{\theta}{2} + \tan \frac{\theta}{2}$$

9. Use the identity:
   $$\cot x + \tan x = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \frac{\cos^2 x + \sin^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$$

So,
   $$E = \frac{1}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}$$

10. Use the double angle identity:
    $$\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \implies \sin \frac{\theta}{2} \cos \frac{\theta}{2} = \frac{\sin \theta}{2}$$

11. Substitute back:
    $$E = \frac{1}{\frac{\sin \theta}{2}} = \frac{2}{\sin \theta} = 2 \csc \theta$$

**Answer:** $2 \csc \theta$

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