Question

cosecθ(cotθ+cosecθ)+cosecθ(tanθ+secθ)=2(1+cosecθ+secθ)

Answer 1

We have reviewed and slightly altered the problem because it was incorrect

$$\operatorname {csc} \theta (\cot \theta + \csc \theta) + \csc \theta (\tan \theta + \sec \theta) = 2 \csc \theta \csc 2 \theta (1 + \sin \theta + \cos \theta)$$

Solution.

Take $\csc \theta$ out of the brackets:

$$\operatorname {csc} \theta (\cot \theta + \csc \theta) + \csc \theta (\tan \theta + \sec \theta) = \csc \theta (\cot \theta + \tan \theta + \csc \theta + \sec \theta)$$

We have

$$\cot \theta + \tan \theta$$

We know that $\cot \theta = \frac{\cos\theta}{\sin\theta}$ and $\tan \theta = \frac{\sin\theta}{\cos\theta}$, then summarize them:

$$\cot \theta + \tan \theta = \frac {\cos^ {2} \theta + \sin^ {2} \theta}{\sin \theta \cos \theta}$$

Use the basic relationship between the sine and the cosine: $\cos^2\theta +\sin^2\theta = 1$

$$\frac {\cos^ {2} \theta + \sin^ {2} \theta}{\sin \theta \cos \theta} = \frac {1}{\sin \theta \cos \theta} = \frac {2}{2 \sin \theta \cos \theta} = \frac {2}{\sin 2 \theta}$$

And at the end

$$\frac {2}{\sin 2 \theta} = 2 \csc 2 \theta = \cot \theta + \tan \theta$$

Similarly, for $\csc \theta +\sec \theta$ :

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

Reduce to a common denominator:

$$\frac {1}{\sin \theta} + \frac {1}{\cos \theta} = \frac {\cos \theta + \sin \theta}{\sin \theta \cos \theta} = (\cos \theta + \sin \theta) \frac {2}{2 \sin \theta \cos \theta} = 2 \csc 2 \theta (\cos \theta + \sin \theta) = \csc \theta + \sec \theta$$

Then

$$\csc \theta (\cot \theta + \tan \theta + \csc \theta + \sec \theta) = 2 \csc \theta \csc 2 \theta (1 + \sin \theta + \cos \theta)$$

We have identity.