+91-8076753736 [email protected]

-->

Inverse Trigonometric Functions

Question

Wired Faculty App

Evaluate

$sin open square brackets 2 space cos to the power of negative 1 end exponent open parentheses negative 3 over 5 close parentheses close square brackets$

Solution

Let space straight y equals cos to the power of negative 1 end exponent open parentheses negative 3 over 5 close parentheses therefore space space space cos space straight y equals negative 3 over 5 space and space straight y element of space left square bracket 0 comma space straight pi right square bracket therefore space space space sin space straight y space is space plus space ve sin space straight y space equals space square root of 1 minus cos squared end root space straight y equals square root of 1 minus open parentheses negative 3 over 5 close parentheses squared end root equals square root of 1 minus 9 over 25 end root equals 4 over 5

Not given expression =

sin open square brackets 2 space cos to the power of negative 1 end exponent open parentheses negative 3 over 5 close parentheses close square brackets equals space s n space left square bracket 2 space y right square bracket space equals space 2 space sin space y space cos space y equals 2 open parentheses 4 over 5 close parentheses open parentheses negative 3 over 5 close parentheses equals negative 24 over 25 space space space space space

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation