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Q:

यदि $$ sec θ ={13 \over 12}$$, तथा θ न्यूनकोण है, तो $$ \sqrt { cotθ+tanθ} \ $$ है....

  • 1
    $$ {4 \over 13}$$
  • 2
    $$ {12 \over 2\sqrt{13}}$$
  • 3
    $$ {13 \over 2\sqrt{15}}$$
  • 4
    $$ {2 \over 13}$$
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Answer : 2. "$$ {12 \over 2\sqrt{13}}$$"

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