Math, asked by moul1998, 1 year ago

limit 1-tanx∕cos 2x
x→π∕4


426ri543: From which class this questions is

Answers

Answered by MADHANSCTS
14
         \lim_{x \to \ \pi /4}             \frac{1-tanx}{cos2x}
     
          \lim_{x \to \ \pi /4}            \frac{1- \frac{sinx}{cosx} }{cos2x}

         \lim_{x \to \ \pi /4}         \frac{ \frac{cosx-sinx}{cosx} }{cos2x}    
           
         \lim_{x \to \ \pi /4}          \frac{ cosx-sinx }{cosx.cos2x}  

         \lim_{x \to \ \pi /4}            \frac{ cosx-sinx }{cosx(cos^{2}x-sin^{2}x)  } 

         \lim_{x \to \ \pi /4}           \frac{1}{cosx(cosx+sinx)}   

                                  =   \frac{1}{ \frac{1}{ \sqrt{2} }.( \frac{1}{ \sqrt{2} }+\frac{1}{ \sqrt{2} })  }

                                 = \frac{1}{ \frac{1}{ \sqrt{2} }. \frac{2}{ \sqrt{2} }  }
                    
                                 = \frac{1}{ \frac{\sqrt{2} }{ \sqrt{2} } }
            
                                 =  \frac{1}{1}

                                 =1





 



Anonymous: what is second step?
MADHANSCTS: tanx = sinx / cosx
Anonymous: not that , it's below step
Anonymous: there u should write cos x- sinx know.........how it became cos 8x?
MADHANSCTS: It is cosx - sinx
MADHANSCTS: look once
MADHANSCTS: it is not cos 8x
Answered by nakshathranambiar200
1

Answer:

1

Step-by-step explanation:

hope it helps you please mark as brainliest and thank you

Similar questions