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Inverse Trigonometric Functions
NCERT Class 12
CBSE
Grade 12
Question:
Prove that $\sin^{-1}\left(\dfrac{3}{5}\right) + \sin^{-1}\left(\dfrac{8}{17}\right) = \cos^{-1}\left(\dfrac{36}{85}\right)$. Also evaluate $\tan\left(\sin^{-1}\dfrac{3}{5} + \sin^{-1}\dfrac{8}{17}\right)$.
Step-by-Step Solution
Part 1: $\sin A = 3/5, \cos A = 4/5$; $\sin B = 8/17, \cos B = 15/17$. [1.0 Mark] $\cos(A+B) = (4/5)(15/17) - (3/5)(8/17) = 36/85 \Rightarrow A+B = \cos^{-1}(36/85)$. Proved! [2.0 Marks] Part 2: $\tan A = 3/4, \tan B = 8/15$. [0.5 Mark] $\tan(A+B) = \dfrac{3/4 + 8/15}{1 - 24/60} = \dfrac{77}{36}$. [1.5 Marks]
--- 🎯 Official CBSE Marking Scheme: Finding trig components for $A$ and $B$: 1.0 Mark Proving $\cos(A+B) = 36/85$: 2.0 Marks Evaluating $\tan(A+B) = 77/36$: 2.0 Marks
Correct Answer:
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