Question

tan-¹ 3/5 + tasln -¹ 1/4 = π​

Answer 1

Correct Question :

$tan^{-1} \frac{3}{5}+tan^{-1} \frac{1}{4}=\frac{\pi}{4}$

Formula used :

$tan^{-1}a+tan^{-1}b=tan^{-1}\frac{a+b}{1-ab}$

Now as per the question,

$tan^{-1} \frac{3}{5}+tan^{-1} \frac{1}{4}$

$=tan^{-1} \frac{\frac{3}{5}+\frac{1}{4}}{1-\frac{3}{20}}\\\\=tan^{-1}1$

= $\frac{\pi}{4}$

Hence Proved...

Answer 2

$\huge \fcolorbox{e}{pink}{Solution :)}$

Given ,

$\large \mathtt{ \fbox{ {Tan}^{ - 1} (\frac{3}{5}) + {Tan}^{ - 1} (\frac{1}{4} ) = \frac{\pi}{4} }}$

We know that ,

$\large \mathtt{\fbox{ {Tan}^{ - 1}(x) + {Tan}^{ - 1} (y) = {Tan}^{ - 1} ( \frac{x + y}{1 - xy} )}}$

$\bigstar$ LHS

$\sf \mapsto {Tan}^{ - 1} (\frac{3}{5}) + {Tan}^{ - 1} (\frac{1}{4} ) = \frac{\pi}{4} \\ \\ \sf \mapsto {Tan}^{ - 1}( \frac{ \frac{3}{5} + \frac{1}{4} }{ 1- \frac{3}{5} \times \frac{1}{4} } ) \\ \\ \sf \mapsto {Tan}^{ - 1}( \frac{ \frac{12 + 5}{20} }{ \frac{20 - 3}{20} } ) \\ \\ \sf \mapsto {Tan}^{ - 1}( \frac{17}{17} ) \\ \\ \sf \mapsto {Tan}^{ - 1}(1) \\ \\ \sf \mapsto \frac{\pi}{4}$

$\bigstar$ RHS

$\therefore \mathtt{LHS = RHS}$

Hence proved

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