please explain the solution of Q:7(iv) in R.D.SHARMA BOOK
 

Asked by mrinalini mrinalini | 21st May, 2014, 10:26: PM

Expert Answer:

T h e space q u e s t i o n space i s : E x. space 5.3 7 left parenthesis i v right parenthesis P r o v e space t h a t space fraction numerator cos 80 degree over denominator sin 10 degree end fraction plus cos 59 degree cross times cos e c 31 degree equals 2 S o l u t i o n : L e t space u s space n o w space c o n s i d e r space t h e space l e f t space h a n d space s i d e space o f space t h e space g i v e n space e x p r e s s i o n : L. H. S equals fraction numerator cos 80 degree over denominator sin 10 degree end fraction plus cos 59 degree cross times cos e c 31 degree equals cos 80 degree cross times cos e c 10 degree plus cos 59 degree cross times cos e c 31 degree equals cos 80 degree cross times s e c open parentheses 90 degree minus 10 degree close parentheses plus cos 59 degree cross times s e c open parentheses 90 degree minus 31 degree close parentheses space space space space space left square bracket sin c e space cos e c open parentheses 90 degree minus theta close parentheses equals s e c theta space a n d space s e c open parentheses 90 degree minus theta close parentheses equals cos e c theta right square bracket equals cos 80 degree cross times s e c 80 degree plus cos 59 degree cross times s e c 59 degree equals 1 plus 1 equals 2 equals R. H. S. S i n c e space L. H. S equals space R. H. S comma space h e n c e space p r o v e d.

Answered by Vimala Ramamurthy | 22nd May, 2014, 09:40: AM