Prove that: $(\sin \theta + \csc \theta)^2 + (\cos \theta + \sec \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta$.
Step-by-Step Solution
Key Concept: LHS $= (\sin^2\theta + \csc^2\theta + 2) + (\cos^2\theta + \sec^2\theta + 2) = (\sin^2\theta + \cos^2\theta) + 4 + \csc^2\theta + \sec^2\theta = 1 + 4 + (1 + \cot^2\theta) + (1 + \tan^2\theta) = 7 + \tan^2\theta + \cot^2\theta$.
LHS $= \sin^2\theta + \csc^2\theta + 2 + \cos^2\theta + \sec^2\theta + 2$. [1.0 Mark]
$= (\sin^2\theta + \cos^2\theta) + 4 + (1 + \cot^2\theta) + (1 + \tan^2\theta)$. [1.0 Mark]
$= 1 + 4 + 1 + \cot^2\theta + 1 + \tan^2\theta = 7 + \tan^2\theta + \cot^2\theta = $ RHS. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Expanding both squared binomials: 1.0 Mark
Applying identities $\sin^2\theta+\cos^2\theta=1$, $\csc^2\theta=1+\cot^2\theta$, $\sec^2\theta=1+\tan^2\theta$: 1.0 Mark
Evaluating sum $= 7 + \tan^2\theta + \cot^2\theta$: 1.0 Mark
Correct Answer: