Question

Prove that $\dfrac{1 + \sin \theta}{\cos \theta} + \dfrac{\cos \theta}{1 + \sin \theta} = 2 \sec \theta$.

Hint:

Use $\sin^2 \theta + \cos^2 \theta = 1$ whenever both $\sin \theta$ and $\cos \theta$ appear together.

Answer 1

Step 1: Take the LHS.
\[ \text{LHS} = \dfrac{1 + \sin \theta}{\cos \theta} + \dfrac{\cos \theta}{1 + \sin \theta} \] Step 2: Take a common denominator.
\[ \text{LHS} = \dfrac{(1 + \sin \theta)^2 + \cos^2 \theta}{\cos \theta (1 + \sin \theta)} \] Step 3: Expand the numerator.
\[ (1 + \sin \theta)^2 + \cos^2 \theta = 1 + 2\sin \theta + \sin^2 \theta + \cos^2 \theta \] \[ = 2(1 + \sin \theta) \] Step 4: Simplify.
\[ \text{LHS} = \dfrac{2(1 + \sin \theta)}{\cos \theta (1 + \sin \theta)} = \dfrac{2}{\cos \theta} = 2 \sec \theta \] Step 5: Conclusion.
Hence proved, \[ \dfrac{1 + \sin \theta}{\cos \theta} + \dfrac{\cos \theta}{1 + \sin \theta} = 2 \sec \theta \]