CBSE Class 12-science Answered

If += $sin to the power of negative 1 end exponent fraction numerator c squared over denominator a b end fraction$ ,prove that +.

Asked by www.nishikadas | 16 May, 2016, 08:54: AM

Expert Answer

$C o n s i d e r space t h e space g i v e n space e q u a t i o n s i n to the power of negative 1 end exponent open parentheses x over a close parentheses plus s i n to the power of negative 1 end exponent open parentheses y over b close parentheses equals s i n to the power of negative 1 end exponent fraction numerator c squared over denominator a b end fraction W e space k n o w space t h a t comma space s i n to the power of negative 1 end exponent x plus s i n to the power of negative 1 end exponent y equals s i n to the power of negative 1 end exponent open parentheses x square root of 1 minus y squared end root plus y square root of 1 minus x squared end root close parentheses T h u s comma s i n to the power of negative 1 end exponent open parentheses x over a close parentheses plus s i n to the power of negative 1 end exponent open parentheses y over b close parentheses equals s i n to the power of negative 1 end exponent open parentheses x over a square root of 1 minus y over b squared squared end root plus y over b square root of 1 minus x over a squared squared end root close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals s i n to the power of negative 1 end exponent open parentheses fraction numerator x over denominator a b end fraction square root of b squared minus y squared end root plus fraction numerator y over denominator a b end fraction square root of a squared minus x squared end root close parentheses B u t space g i v e n space t h a t comma space s i n to the power of negative 1 end exponent open parentheses x over a close parentheses plus s i n to the power of negative 1 end exponent open parentheses y over b close parentheses equals s i n to the power of negative 1 end exponent fraction numerator c squared over denominator a b end fraction therefore s i n to the power of negative 1 end exponent open parentheses fraction numerator x over denominator a b end fraction square root of b squared minus y squared end root plus fraction numerator y over denominator a b end fraction square root of a squared minus x squared end root close parentheses equals s i n to the power of negative 1 end exponent fraction numerator c squared over denominator a b end fraction rightwards double arrow fraction numerator x over denominator a b end fraction square root of b squared minus y squared end root plus fraction numerator y over denominator a b end fraction square root of a squared minus x squared end root equals fraction numerator c squared over denominator a b end fraction rightwards double arrow x square root of b squared minus y squared end root plus y square root of a squared minus x squared end root equals c squared rightwards double arrow open parentheses x square root of b squared minus y squared end root plus y square root of a squared minus x squared end root close parentheses squared equals open parentheses c squared close parentheses squared rightwards double arrow x squared open parentheses b squared minus y squared close parentheses plus y squared open parentheses a squared minus x squared close parentheses plus 2 x y square root of b squared minus y squared end root square root of a squared minus x squared end root equals c to the power of 4 P l e a s e space c h e c k space t h e space q u e s t i o n.$