# How to solve $\frac{\sin\theta+\cos\theta}{\sec\theta+\csc\theta} = 1 /\sqrt8$

$$\frac{\sin\theta+\cos\theta}{\sec\theta+\csc\theta} = 1/ \sqrt8$$

Here is my steps

$$\frac{\sin\theta+\cos\theta}{\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}} = 1/√8$$

$$\sin\theta+\cos\theta * \frac{\sin\theta\cos\theta}{\sin\theta+\cos\theta} = 1/√8$$

And then I multiplied sin+cos with denominator to get same denominator and after this,

$$\frac{\sin^2\theta+(2)\sin\theta\cos\theta+\cos^2\theta*\sin\theta\cos\theta}{\sin\theta+\cos\theta} = 1/√8$$

How to carry on?

• Can you show the intermediate steps? I think you might have made a mistake Mar 11 at 21:31
• @CalvinLin ok but I will take time to write it Mar 11 at 21:32
• You have a division slash in the first line that is not in the title, nor in the last line. I am sure it should be in the title, but do not know if you inverted things along the way. Please clarify. Mar 11 at 21:35
• The problem is that you turned $$\frac{\sin \theta + \cos \theta}{\frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta}}$$ into $$(\sin \theta + \cos \theta)\frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta}.$$ These are not the same. Mar 11 at 22:00

Note that $$\sec \theta + \csc \theta = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta}.$$

Thus, $$\frac{\sin \theta + \cos \theta}{\sec \theta + \csc \theta} = \frac{\sin \theta + \cos \theta}{\frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta}}.$$

This simplifies to something much nicer than the expression you gave. Can you take it from here?

• I already reached here and continued after this Mar 11 at 21:51
• I edited the question, please check Mar 11 at 21:56

As argued in diracdeltafunk's post,

$$\frac{\sin\theta+\cos\theta}{\sec\theta+\csc\theta}= \frac{\sin\theta+\cos\theta}{\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}}=\sin\theta\cos\theta.$$

Now the double angle formula reads $$\sin(2\theta)=2\sin\theta\cos\theta$$ which yields the equivalent formulation of the problem $$\sin(2\theta)=\frac{1}{\sqrt{2}}.$$ From this, one can easily deduct $$\theta \in \bigg\{\frac{\pi}{8}+\pi k : k\in \mathbb{N}\bigg\}\cup \bigg\{\frac{3\pi}{8}+\pi k : k\in \mathbb{N}\bigg\}.$$