+91-8076753736[email protected]
logo
logo
-->

Inverse Trigonometric Functions

Question
CBSEENMA12032628
Wired Faculty App

Find the principal value of sin–1  open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses.

Solution

 Let space space straight y space equals sin to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses space where space space straight pi over 2 less or equal than space straight y space less or equal than space straight pi over 2 therefore space space space sin space straight y space equals fraction numerator 1 over denominator square root of 2 end fraction space where space minus straight pi over 2 space less or equal than space straight y space less or equal than space straight pi over 2 therefore space space space space space space space space straight y equals straight pi over 4 therefore space space space space space space space required space principal space value space equals space straight pi over 4

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation