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Inverse Trigonometric Functions

Question
CBSEENMA12036032
Wired Faculty App

straight l im with straight x rightwards arrow 0 below space fraction numerator left parenthesis 1 minus cos 2 straight x right parenthesis left parenthesis 3 space plus cos space straight x right parenthesis over denominator straight x space tan space 4 straight x end fraction space is equal to 

  • 4

  • 3

  • 2

  • 1/2

Solution

C.

2

We have,
limit as straight x space rightwards arrow 0 of space fraction numerator left parenthesis 1 minus cos space 2 straight x right parenthesis left parenthesis 3 plus cosx over denominator straight x space tan space 4 straight x end fraction space equals limit as straight x space rightwards arrow 0 of fraction numerator 2 sin squared space straight x space left parenthesis 3 plus cosx right parenthesis over denominator straight X space straight x space begin display style fraction numerator tanx over denominator 4 straight x end fraction end style space straight x space 4 straight x end fraction equals limit as straight x space rightwards arrow 0 of fraction numerator 2 sin squared straight x over denominator straight x squared end fraction space straight x space limit as straight x space rightwards arrow 0 of fraction numerator 3 plus cos space straight x over denominator 4 end fraction space straight x limit as straight x space rightwards arrow 0 of fraction numerator 1 over denominator begin display style fraction numerator tan begin display style space end style begin display style 4 end style begin display style straight x end style over denominator 4 straight x end fraction end style end fraction space equals space 2 space straight x space 4 over 4 space straight x space 1 space equals space 2

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