Please solve the following question:
Find general formula of the given equation-
 
open parentheses table row cell square root of 3 minus 1 end cell end table close parentheses cos theta plus open parentheses table row cell square root of 3 plus 1 end cell end table close parentheses sin theta equals 2

Asked by manasd | 26th Nov, 2015, 10:06: PM

Expert Answer:

begin mathsize 14px style open parentheses square root of 3 minus 1 close parentheses cos theta plus open parentheses square root of 3 plus 1 close parentheses sin theta equals 2 D i v i d e space b o t h space s i d e s space b y space 2 square root of 2 fraction numerator open parentheses square root of 3 minus 1 close parentheses over denominator 2 square root of 2 end fraction cos theta plus fraction numerator open parentheses square root of 3 plus 1 close parentheses over denominator 2 square root of 2 end fraction sin theta equals fraction numerator 1 over denominator 2 square root of 2 end fraction sin open square brackets theta plus tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 3 minus 1 over denominator square root of 3 plus 1 end fraction close parentheses close square brackets equals sin straight pi over 4 theta equals 2 n straight pi plus straight pi over 3 space space or space space 2 nπ minus straight pi over 6 space space space straight n element of straight Z end style

Answered by Rashmi Khot | 27th Nov, 2015, 08:49: AM