-->

Inverse Trigonometric Functions

Question
CBSEENMA12032656

Find the value of tan-1 open square brackets 2 space cos space open parentheses 2 space sin to the power of negative 1 end exponent 1 half close parentheses close square brackets

Solution
Let space space space space straight y space equals space sin to the power of negative 1 end exponent 1 half space space space space space space where space space minus straight pi over 2 less or equal than straight r less or equal than straight pi over 2
therefore space space space space space space sin space space straight y space equals space 1 half space space space space space space where space space minus straight pi over 2 less or equal than straight r less or equal than straight pi over 2
therefore space space space space space space space straight y space equals straight pi over 6
therefore space space space space space space space tan to the power of negative 1 end exponent open square brackets 2 space cos space open parentheses 2 space sin to the power of negative 1 end exponent space 1 half close parentheses close square brackets equals space tan to the power of negative 1 end exponent space open square brackets 2 space cos space open parentheses 2 cross times straight pi over 6 close parentheses close square brackets
space space space space space space space space space space space space tan space to the power of negative 1 end exponent open square brackets 2 space cos straight pi over 3 close square brackets equals tan to the power of negative 1 end exponent open square brackets 2 cross times 1 half close square brackets equals tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis equals straight pi over 4

Some More Questions From Inverse Trigonometric Functions Chapter