Question

Prove that $(\sin \theta + \csc \theta)^2 + (\cos \theta + \sec \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta$.

Hint:

Always expand the given trigonometric expression using basic identities like $\sin \theta \csc \theta = 1$ and $\sin^2 \theta + \cos^2 \theta = 1$. Simplify step by step.

Answer 1

Step 1: Start with the left-hand side (L.H.S).
\[ \text{L.H.S.} = (\sin \theta + \csc \theta)^2 + (\cos \theta + \sec \theta)^2 \] Expand both terms: \[ = (\sin^2 \theta + \csc^2 \theta + 2\sin \theta \csc \theta) + (\cos^2 \theta + \sec^2 \theta + 2\cos \theta \sec \theta) \] Step 2: Simplify using trigonometric identities.
\[ \sin \theta \csc \theta = 1, \quad \cos \theta \sec \theta = 1 \] So, \[ \text{L.H.S.} = \sin^2 \theta + \cos^2 \theta + \csc^2 \theta + \sec^2 \theta + 2 + 2 \] \[ \Rightarrow \text{L.H.S.} = (\sin^2 \theta + \cos^2 \theta) + (\csc^2 \theta + \sec^2 \theta) + 4 \] Step 3: Use the fundamental identity.
\[ \sin^2 \theta + \cos^2 \theta = 1 \] \[ \Rightarrow \text{L.H.S.} = 1 + (\csc^2 \theta + \sec^2 \theta) + 4 = 5 + \csc^2 \theta + \sec^2 \theta \]
Step 4: Express $\csc^2 \theta$ and $\sec^2 \theta$ in terms of $\tan^2 \theta$ and $\cot^2 \theta$.
\[ \csc^2 \theta = 1 + \cot^2 \theta, \quad \sec^2 \theta = 1 + \tan^2 \theta \] Substitute these: \[ \text{L.H.S.} = 5 + (1 + \cot^2 \theta) + (1 + \tan^2 \theta) \] \[ \text{L.H.S.} = 7 + \tan^2 \theta + \cot^2 \theta \]
Step 5: Therefore, \[ \boxed{\text{L.H.S.} = \text{R.H.S.}} \] Hence proved.