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`        maximum value of the expression 1/sin^2theta+3sinthetacostheta+5cos^2theta `
one year ago

## Answers : (2)

```							For the given expression to be maximum , denominator should be minimum which is sin^2x+3sinxcosx+5 cos^2x should be minimum.let y= sin^2x+3sinxcosx+5cos^2x2y= 2sin^2x+6sinxcosx+10cos^2x(I am multiplying by 2 on both sides for simple calculation . )2y =sin^2x + 2 (sin x) (3cosx) + (3 cos x )^2 + sin^2x+cos^2x2y= (sin x + 3 cos x ) ^2 +1The minimum value of (sin x+3 cos x )^2 is 0since sin x + 3cos x can be expressed as a single sine function,sin (x+ n ) where n = arcsin ( 3/ sqrt10)so the minimum value of 2y is 0+1 =1and that of y is 1/2so the minimum value of sin^2 x + 3sinxcosx + cos^2x is 0.5the maximum value of 1/(sin^2 x + 3sinxcosx + cos^2x) is 2
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one year ago
```							dear student.the maximum and minimum value accoding to the above given question is 2 and 0.5 respectively .
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one year ago
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