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Q.26 answer for 2 nd or question alone, answer as soon as possible.

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8 months ago

```							call theta= ax= e^a(2cosa+cos2a) so that dx/da= e^a(2cosa+cos2a – 2sina – 2sin2a)y= e^a(2sina+sin2a) so that dy/da= e^a(2sina+sin2a+2cosa+2cos2a)now, by chain rule dy/dx= [dy/da]/[dx/da]= e^a(2sina+sin2a+2cosa+2cos2a)/e^a(2cosa+cos2a – 2sina – 2sin2a)= (2sina+sin2a+2cosa+2cos2a)/(2cosa+cos2a – 2sina – 2sin2a), which is clearly not equal to tan(pi/4+3a/2), as can be seen by putting a= 0, whence (2sina+sin2a+2cosa+2cos2a)/(2cosa+cos2a – 2sina – 2sin2a)= 4/3 but tan(pi/4+3a/2)= 1.hence question is certainly WRONG, as can also be verified on wolfram alpha ‘paramteric derivative calculator’KINDLY APPROVE :))
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8 months ago
```							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)  = [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 Thanks and regardsArun
```
8 months ago
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