+91-8076753736 [email protected]

-->

Inverse Trigonometric Functions

Question

Wired Faculty App

Simplify

$cot to the power of negative 1 end exponent open parentheses square root of 1 plus straight x squared end root plus straight x close parentheses$

Solution

Let space space space space straight z equals cot to the power of negative 1 end exponent open parentheses square root of 1 plus straight x squared end root plus straight x close parentheses Put space straight x equals tan space straight theta therefore space space space straight z equals cot to the power of negative 1 end exponent open parentheses square root of 1 plus tan squared space straight theta end root plus tan space straight theta close parentheses equals cot to the power of negative 1 end exponent space space equals cot to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator cos space straight theta end fraction plus fraction numerator sin space straight theta over denominator cos space straight theta end fraction close parentheses equals cot to the power of negative 1 end exponent open parentheses fraction numerator 1 plus sin space straight theta over denominator cos space straight theta end fraction close parentheses space space equals cot to the power of negative 1 end exponent open parentheses fraction numerator cos squared begin display style straight theta over 2 end style plus sin squared begin display style straight theta over 2 end style plus 2 space cos begin display style straight theta over 2 end style sin begin display style straight theta over 2 end style over denominator cos squared begin display style straight theta over 2 end style minus sin squared begin display style straight theta over 2 end style end fraction close parentheses equals cot to the power of negative 1 end exponent open square brackets fraction numerator open parentheses cos begin display style straight theta over 2 end style plus sin begin display style straight theta over 2 end style close parentheses squared over denominator open parentheses cos begin display style straight theta over 2 end style plus sin begin display style straight theta over 2 end style close parentheses open parentheses cos straight theta over 2 minus sin straight theta over 2 close parentheses end fraction close square brackets space equals space cot to the power of negative 1 end exponent open square brackets fraction numerator cos begin display style straight theta over 2 end style plus sin begin display style straight theta over 2 end style over denominator cos begin display style straight theta over 2 end style minus sin begin display style straight theta over 2 end style end fraction close square brackets equals cot to the power of negative 1 end exponent open square brackets fraction numerator 1 plus tan begin display style straight theta over 2 end style over denominator 1 minus tan begin display style straight theta over 2 end style end fraction close square brackets equals cot to the power of negative 1 end exponent open square brackets cot open parentheses straight pi over 4 minus straight theta over 2 close parentheses close square brackets equals straight pi over 4 minus straight theta over 2 equals straight pi over 4 minus 1 half tan to the power of negative 1 end exponent space straight x

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation