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`        Q.26 second OR questionanswer as soon as possible.`
3 months 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
```
3 months 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
```
3 months ago
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