If 0 < theta < pi/2 and if (y+1)/(1-y) = square root of blank end root1 +sintheta/1-sintheta then y =
 
a) cot theta/2 
b) tantheta/2
c) cot theta/2 + tan theta/2
d) cottheta/2 - tan theta/2

Asked by ahuja8087 | 13th Sep, 2015, 11:13: PM

Expert Answer:

G i v e n space t h a t fraction numerator 1 plus y over denominator 1 minus y end fraction equals square root of fraction numerator 1 plus sin theta over denominator 1 minus sin theta end fraction end root R a t i o n a l i sin g space t h e space d e n o m i n a t o r space o n space t h e space l e f t space h a n d space s i d e comma space w e space h a v e comma fraction numerator 1 plus y over denominator 1 minus y end fraction equals square root of fraction numerator 1 plus s i n theta over denominator 1 minus s i n theta end fraction cross times fraction numerator 1 plus s i n theta over denominator 1 plus s i n theta end fraction end root rightwards double arrow fraction numerator 1 plus y over denominator 1 minus y end fraction equals square root of fraction numerator open parentheses 1 plus sin theta close parentheses squared over denominator 1 minus s i n squared theta end fraction end root rightwards double arrow fraction numerator 1 plus y over denominator 1 minus y end fraction equals square root of fraction numerator open parentheses 1 plus s i n theta close parentheses squared over denominator cos squared theta end fraction end root rightwards double arrow fraction numerator 1 plus y over denominator 1 minus y end fraction equals fraction numerator open parentheses 1 plus s i n theta close parentheses over denominator c o s theta end fraction A p p l y i n g space c o m p o n e n d o space a n d space d i v i d e n d o space o n space b o t h space t h e space s i d e s comma space w e space h a v e comma fraction numerator 1 plus y plus 1 minus y over denominator 1 plus y minus 1 plus y end fraction equals fraction numerator 1 plus s i n theta plus c o s theta over denominator 1 plus s i n theta minus c o s theta end fraction rightwards double arrow fraction numerator 2 over denominator 2 y end fraction equals fraction numerator 1 plus s i n theta plus c o s theta over denominator 1 plus s i n theta minus c o s theta end fraction rightwards double arrow y equals fraction numerator 1 plus s i n theta minus c o s theta over denominator 1 plus s i n theta plus c o s theta end fraction rightwards double arrow y equals fraction numerator 2 s i n begin display style theta over 2 end style cos theta over 2 plus 2 s i n squared theta over 2 over denominator 2 s i n theta over 2 c o s theta over 2 plus 2 cos squared theta over 2 end fraction rightwards double arrow y equals fraction numerator 2 s i n begin display style theta over 2 end style open square brackets cos theta over 2 plus sin theta over 2 close square brackets over denominator 2 c o s theta over 2 open square brackets c o s theta over 2 plus s i n theta over 2 close square brackets end fraction rightwards double arrow y equals tan theta over 2 H e n c e space t h e space c o r r e c t space o p t i o n space i s space left parenthesis b right parenthesis.

Answered by Vimala Ramamurthy | 14th Sep, 2015, 12:10: PM