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Prove that ^{-1}{1}{4}+^{-1}{2}{9}=^{-1}{1}{{5}}

Question

Prove that $tan^{- 1} \frac{1}{4} + tan^{- 1} \frac{2}{9} = sin^{- 1} \frac{1}{5}$

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Prove that

tan^{- 1} \frac{1}{4} + tan^{- 1} \frac{2}{9} = sin^{- 1} \frac{1}{5}

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Text solutionVerified

To prove:

tan^{- 1} \frac{1}{4} + tan^{- 1} \frac{2}{9} = sin^{- 1} \frac{1}{5}

Proof:

Let

tan^{- 1} \frac{1}{4} = x \Rightarrow tan x = \frac{1}{4}

\Rightarrow cos x = \frac{4}{17}

\Rightarrow sin x = \frac{1}{17}

Again, let

tan^{- 1} \frac{2}{9} = y \Rightarrow tan y = \frac{2}{9}

\Rightarrow cos y = \frac{9}{85}

\Rightarrow sin y = \frac{2}{85}

We know that

sin (x + y) = sin x . cos y + cos x . sin y

= \frac{1}{17} . \frac{9}{85} + \frac{4}{17} . \frac{2}{85}

= \frac{17}{17 . 85} = \frac{17}{17 . 5} = \frac{1}{5}

\Rightarrow (x + y) = sin^{- 1} \frac{1}{5}

\Rightarrow tan^{- 1} \frac{1}{4} + tan^{- 1} \frac{2}{9} = sin^{- 1} \frac{1}{5}

Hence proved.

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Question Text	Prove that $tan^{- 1} \frac{1}{4} + tan^{- 1} \frac{2}{9} = sin^{- 1} \frac{1}{5}$
Answer Type	Text solution:1
Upvotes	150