Question

tan-1(3sin2x/5+3cos2x) + tan-1(1/4tanx)

Answer 1

^_^ INVERSE TRIGONOMETRY ^.^

By Looking at other sources :
Correct question : 
Question:  Prove that 
$tan^{-1} ( \frac{3sin(2x)}{5+3cos(2x)} ) \:\: + tan^{-1}( \frac{tan(x)}{4} ) = x$

Solution: 
Steps:
1) We know that , 
    $sin(2x) = \frac{2 tan(x)}{1+ tan^{2}(x) } \\ cos(2x) = \frac{1- tan^{2}(x) }{1+ tan^{2}(x) }$

2)  Substitute above values of sin(2x) and cos(2x)  in LHS .

 LHS =  $tan^{-1} ( \frac{3sin(2x)}{5+3cos(2x)} ) \:\: + tan^{-1}( \frac{tan(x)}{4} )$
 
          =  $tan^{-1} ( \frac{3sin(2x)}{5+3cos(2x)} ) + \:\: tan^{-1}( \frac{tan(x)}{4} ) \\ \\=\ \textgreater \ tan^{-1} ( \frac{3* \frac{2tan(x)}{1+ tan^{2}(x) } }{5+3( \frac{1- tan^{2}(x) }{1+ tan^{2}(x) } )} ) + \:\: tan^{-1}( \frac{tan(x)}{4} )\\=\ \textgreater \ tan^{-1} ( \frac{6tan(x)}{8+2 tan^{2}(x) } ) + \:\: tan^{-1}( \frac{tan(x)}{4} )\\ \\ =\ \textgreater \ tan^{-1} ( \frac{ \frac{3}{4} tan(x)}{1+ \frac{1}{4} tan^{2}(x) } ) + \:\: tan^{-1}( \frac{tan(x)}{4} )$     

$tan^{-1} ( \frac{tan(x)- \frac{1}{4} tan(x)}{1+ tan(x) * \frac{1}{4} tan(x) } ) + \:\: tan^{-1}( \frac{tan(x)}{4} ) \\ \\ =\ \textgreater \ [tan^{-1} (tan(x)) - tan^{-1} ( \frac{tan(x)}{4} ) ] \;+ tan^{-1} ( \frac{tan(x)}{4} ) \\ =\ \textgreater \ x$ 

Hence Proved 

Answer 2

Answer:

Step-by-step explanation:

LHS =

tan^{-1} ( \frac{3sin(2x)}{5+3cos(2x)} ) \:\: + tan^{-1}( \frac{tan(x)}{4} )

         =  tan^{-1} ( \frac{3sin(2x)}{5+3cos(2x)} ) + \:\: tan^{-1}( \frac{tan(x)}{4} ) \\ \\=\ \textgreater \ tan^{-1} ( \frac{3* \frac{2tan(x)}{1+ tan^{2}(x) } }{5+3( \frac{1- tan^{2}(x) }{1+ tan^{2}(x) } )} ) + \:\: tan^{-1}( \frac{tan(x)}{4} )\\=\ \textgreater \ tan^{-1} ( \frac{6tan(x)}{8+2 tan^{2}(x) } ) + \:\: tan^{-1}( \frac{tan(x)}{4} )\\ \\ =\ \textgreater \ tan^{-1} ( \frac{ \frac{3}{4} tan(x)}{1+ \frac{1}{4} tan^{2}(x) } ) + \:\: tan^{-1}( \frac{tan(x)}{4} ) \\ \\      

tan^{-1} ( \frac{tan(x)- \frac{1}{4} tan(x)}{1+ tan(x) * \frac{1}{4} tan(x) } ) + \:\: tan^{-1}( \frac{tan(x)}{4} ) \\ \\ =\ \textgreater \ [tan^{-1} (tan(x)) - tan^{-1} ( \frac{tan(x)}{4} ) ] \;+ tan^{-1} ( \frac{tan(x)}{4} ) \\ =\ \textgreater \ x