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Q.26 second OR question answer as soon as possible.

Q.26 second OR question
answer as soon as possible.

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Grade:12

2 Answers

Arun
25763 Points
2 years ago
Dear student
 
I have assumed theta as a for the sake of writing
 
dy/da = e^a ( 2 sin a + sin2a) + e^a ( 2 cos a + 2 cos2a)
 
dx/da = e^a ( 2 cosa + cos2a) -  e^a ( 2 sin a - 2 sin2a)
 
Hence 
dy/dx = (sina  + sin2a + cos a + cos2a) /(cosa + cos2a - sina - sin2a)
 
 = Now use sin2a = 2 sina cosa
 
Arun
25763 Points
2 years ago
dy/da = e^a ( 2 sin a + sin2a) + e^a ( 2 cos a + 2 cos2a)
 
dx/da = e^a ( 2 cosa + cos2a) - e^a ( 2 sin a - 2 sin2a)
 
Hence 
dy/dx = (sina + sin2a + cos a + cos2a) /(cosa + cos2a - sina - sin2a)
 
 = [2 sin (3a/2) cos (a/2) + 2 cos (3a/2) cos (a/2)] / [ 2 cos(3a/2) cos (a/2) - 2 sin (3a/2) cos (a/2)]
 
 = [ sin (3a/2) + cos(3a/2) ]/ [ cos(3a/2) - sin (3a/2)]
 
 =[ 1 + tan (3a/2) ]/ [1 - tan (3a/2)]
 
 = tan (π/4 + 3a/2)
 
Hope it helps

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