Question
Find the positive integral solution of:
\[ \tan^{-1}x + \cos^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) \]
Solution
We know identity:
\[ \cos^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) = \tan^{-1}\left(\frac{1}{y}\right) \]
So equation becomes:
\[ \tan^{-1}x + \tan^{-1}\left(\frac{1}{y}\right) = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) \]
Now,
\[ \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) = \tan^{-1}(3) \]
(since \( \sin\theta = 3/\sqrt{10} \Rightarrow \tan\theta = 3 \))
Thus,
\[ \tan^{-1}x + \tan^{-1}\left(\frac{1}{y}\right) = \tan^{-1}(3) \]
Using identity:
\[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]
So,
\[ \frac{x + \frac{1}{y}}{1 – \frac{x}{y}} = 3 \]
Simplify:
\[ \frac{xy + 1}{y – x} = 3 \]
\[ xy + 1 = 3y – 3x \]
\[ xy + 3x – 3y + 1 = 0 \]
Try positive integers:
\[ (x, y) = (1, 1) \Rightarrow 1 + 3 – 3 + 1 = 2 \neq 0 \]
\[ (x, y) = (1, 2) \Rightarrow 2 + 3 – 6 + 1 = 0 \]
Hence,
\[ x = 1,\quad y = 2 \]
Final Answer:
\[ \boxed{(x, y) = (1, 2)} \]
Key Concept
Convert inverse cosine into inverse tangent and use tangent addition identity.