Question

plz ans second part fast

Answer 1

Hi there!
Here's the answer :

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$tan^{-1}\: (\frac{1}{4})+ tan^{-1}\: (\frac{2}{9})= \frac{1}{2} Cos^{-1}\: (\frac{3}{5})= \frac{1}{2}\: sin^{-1}\: (\frac{4}{5})$

$LHS=$

$tan^{-1}\: (\frac{1}{4})+ tan^{-1}\: (\frac{2}{9})$

Using the Formula,
$tan (A + B) = \frac{tanA + tanB}{1-tanAtanB}$

=> $tan^{-1}\: A + tan^{-1}\: B = tan^{-1}\: \frac{A+B}{1-AB}$

$tan^{-1}\: (\dfrac{\frac{1}{4}+\frac{2}{9}}{1-(\frac{1}{4}×\frac{2}{9})})$

$tan^{-1}\: (\dfrac{\frac{17}{36}}{\frac{34}{36}})$

$tan^{-1}\: (\frac{17}{34})$

$tan^{-1}\: (\frac{1}{2})$

Multiply and Divide with 2

$\frac{2}{2} × tan^{-1}\: (\frac{1}{2})$

$\frac{1}{2} × 2tan^{-1}\: (\frac{1}{2})$

Using the Formula
$2tan^{-1}\: x = cos^{-1}\: (\dfrac{1-\frac{x}{2}}{1+\frac{x}{2}})$

$\dfrac{1}{2} × cos^{-1}\: (\dfrac{1-\frac{1}{4}}{1+\frac{1}{4}})$

$\dfrac{1}{2} × cos^{-1}\: (\dfrac{3}{5})$

3,4&5 are Pythagorean triplets
°•° 5² = 3² + 4², according to Pythagoras theorem

$Cos = \frac{Adjacent}{Hypotenuse}$

$Sin = \frac{Opposite}{Hypotenuse}$

Here,
Adj = 3, Hyp = 5
So Opp = 4

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$\dfrac{1}{2}× sin^{-1}\: (\dfrac{4}{5})$

$=RHS$

Hence Proved.

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