Can anyone Please solve Q 170?
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We will use the following identities:

![\left[\frac{\cos^2\theta}{1+\sin\theta}-\frac{\sin^2\theta}{1+\cos\theta}\right]^2 \\ \\ \\ =\left[\frac{1-\sin^2\theta}{1+\sin\theta} - \frac{1-\cos^2\theta}{1+\cos\theta}\right]^2 \\ \\ \\ = \left[\frac{\cancel{(1+\sin \theta)}(1-\sin\theta)}{\cancel{(1+\sin\theta)}} - \frac{\cancel{(1+\cos\theta)}(1-\cos\theta)}{\cancel{(1+\cos\theta)}}\right]^2 \\ \\ \\ = \left[ 1-\sin \theta - (1-\cos\theta)\right]^2 \\ \\ \\ = \left[1-\sin\theta-1+\cos\theta\right]^2 \\ \\ \\ = \left[\cos \theta-\sin\theta\right]^2 \left[\frac{\cos^2\theta}{1+\sin\theta}-\frac{\sin^2\theta}{1+\cos\theta}\right]^2 \\ \\ \\ =\left[\frac{1-\sin^2\theta}{1+\sin\theta} - \frac{1-\cos^2\theta}{1+\cos\theta}\right]^2 \\ \\ \\ = \left[\frac{\cancel{(1+\sin \theta)}(1-\sin\theta)}{\cancel{(1+\sin\theta)}} - \frac{\cancel{(1+\cos\theta)}(1-\cos\theta)}{\cancel{(1+\cos\theta)}}\right]^2 \\ \\ \\ = \left[ 1-\sin \theta - (1-\cos\theta)\right]^2 \\ \\ \\ = \left[1-\sin\theta-1+\cos\theta\right]^2 \\ \\ \\ = \left[\cos \theta-\sin\theta\right]^2](https://tex.z-dn.net/?f=%5Cleft%5B%5Cfrac%7B%5Ccos%5E2%5Ctheta%7D%7B1%2B%5Csin%5Ctheta%7D-%5Cfrac%7B%5Csin%5E2%5Ctheta%7D%7B1%2B%5Ccos%5Ctheta%7D%5Cright%5D%5E2+%5C%5C+%5C%5C+%5C%5C+%3D%5Cleft%5B%5Cfrac%7B1-%5Csin%5E2%5Ctheta%7D%7B1%2B%5Csin%5Ctheta%7D+-+%5Cfrac%7B1-%5Ccos%5E2%5Ctheta%7D%7B1%2B%5Ccos%5Ctheta%7D%5Cright%5D%5E2+%5C%5C+%5C%5C+%5C%5C+%3D+%5Cleft%5B%5Cfrac%7B%5Ccancel%7B%281%2B%5Csin+%5Ctheta%29%7D%281-%5Csin%5Ctheta%29%7D%7B%5Ccancel%7B%281%2B%5Csin%5Ctheta%29%7D%7D+-+%5Cfrac%7B%5Ccancel%7B%281%2B%5Ccos%5Ctheta%29%7D%281-%5Ccos%5Ctheta%29%7D%7B%5Ccancel%7B%281%2B%5Ccos%5Ctheta%29%7D%7D%5Cright%5D%5E2+%5C%5C+%5C%5C+%5C%5C+%3D+%5Cleft%5B+1-%5Csin+%5Ctheta+-+%281-%5Ccos%5Ctheta%29%5Cright%5D%5E2+%5C%5C+%5C%5C+%5C%5C+%3D+%5Cleft%5B1-%5Csin%5Ctheta-1%2B%5Ccos%5Ctheta%5Cright%5D%5E2+%5C%5C+%5C%5C+%5C%5C+%3D+%5Cleft%5B%5Ccos+%5Ctheta-%5Csin%5Ctheta%5Cright%5D%5E2+)
![= \cos^2\theta -2\cos \theta \sin \theta + \sin^2\theta \\ \\ \\ = (\cos^2\theta+\sin^2\theta)-\sin 2\theta \\ \\ \\ = 1-\sin2\theta \\ \\ \\ \\ \implies \boxed{\left[\frac{\cos^2\theta}{1+\sin\theta}-\frac{\sin^2\theta}{1+\cos\theta}\right]^2=1-\sin 2\theta} = \cos^2\theta -2\cos \theta \sin \theta + \sin^2\theta \\ \\ \\ = (\cos^2\theta+\sin^2\theta)-\sin 2\theta \\ \\ \\ = 1-\sin2\theta \\ \\ \\ \\ \implies \boxed{\left[\frac{\cos^2\theta}{1+\sin\theta}-\frac{\sin^2\theta}{1+\cos\theta}\right]^2=1-\sin 2\theta}](https://tex.z-dn.net/?f=%3D+%5Ccos%5E2%5Ctheta+-2%5Ccos+%5Ctheta+%5Csin+%5Ctheta+%2B+%5Csin%5E2%5Ctheta+%5C%5C+%5C%5C+%5C%5C+%3D+%28%5Ccos%5E2%5Ctheta%2B%5Csin%5E2%5Ctheta%29-%5Csin+2%5Ctheta+%5C%5C+%5C%5C+%5C%5C+%3D+1-%5Csin2%5Ctheta+%5C%5C+%5C%5C+%5C%5C+%5C%5C+%5Cimplies+%5Cboxed%7B%5Cleft%5B%5Cfrac%7B%5Ccos%5E2%5Ctheta%7D%7B1%2B%5Csin%5Ctheta%7D-%5Cfrac%7B%5Csin%5E2%5Ctheta%7D%7B1%2B%5Ccos%5Ctheta%7D%5Cright%5D%5E2%3D1-%5Csin+2%5Ctheta%7D)
Thus, the Answer is Option 2)
Thus, the Answer is Option 2)
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