Inverse Trigonometric Functions

Question

Prove that $sin to the power of negative 1 end exponent x minus sin to the power of negative 1 end exponent y equals sin to the power of negative 1 end exponent open square brackets x square root of 1 minus y squared end root minus y square root of 1 minus x squared end root close square brackets$

Solution

Put $sin to the power of negative 1 end exponent straight x equals straight theta comma space sin to the power of negative 1 end exponent straight y equals straight ϕ$
$therefore space space space space space space straight x equals sinθ comma space straight y equals sinϕ Now space sin space left parenthesis straight theta minus straight ϕ right parenthesis equals sin space straight theta space cos space straight ϕ minus cosθ space sin space straight ϕ therefore space space space space space sin left parenthesis straight theta minus straight ϕ right parenthesis space sin space straight theta square root of 1 minus sin squared straight ϕ end root minus sinϕ square root of 1 minus sin squared straight ϕ end root rightwards double arrow space space space space left parenthesis straight theta minus straight ϕ right parenthesis space equals sin to the power of 1 open square brackets sinθ square root of 1 minus sin squared straight ϕ end root minus sinϕ square root of 1 minus sin squared straight theta end root close square brackets rightwards double arrow space space space space sin to the power of negative 1 end exponent straight x minus sin to the power of negative 1 end exponent straight y equals sin to the power of negative 1 end exponent open square brackets straight x square root of 1 minus straight y squared end root minus straight y square root of 1 minus straight x squared end root close square brackets$

For Daily Free Study Material Join wiredfaculty

Inverse Trigonometric Functions

Prove that $sin to the power of negative 1 end exponent x minus sin to the power of negative 1 end exponent y equals sin to the power of negative 1 end exponent open square brackets x square root of 1 minus y squared end root minus y square root of 1 minus x squared end root close square brackets$

Some More Questions From Inverse Trigonometric Functions Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction