Question

Can anyone Please solve Q 170?

Answer 1

We will use the following identities:


$\rightarrow \cos^2\theta = 1-\sin^2\theta \\ \\ \rightarrow \sin^2\theta = 1-\cos^2\theta \\ \\ \rightarrow a^2-b^2=(a+b)(a-b) \\ \\ \rightarrow (a-b)^2=a^2-2ab+b^2\\ \\ \rightarrow \cos^2\theta + \sin^2\theta=1 \\ \\ \rightarrow 2\sin\theta \cos\theta = \sin 2\theta$



$\left[\frac{\cos^2\theta}{1+\sin\theta}-\frac{\sin^2\theta}{1+\cos\theta}\right]^2 \\ \\ \\ =\left[\frac{1-\sin^2\theta}{1+\sin\theta} - \frac{1-\cos^2\theta}{1+\cos\theta}\right]^2 \\ \\ \\ = \left[\frac{\cancel{(1+\sin \theta)}(1-\sin\theta)}{\cancel{(1+\sin\theta)}} - \frac{\cancel{(1+\cos\theta)}(1-\cos\theta)}{\cancel{(1+\cos\theta)}}\right]^2 \\ \\ \\ = \left[ 1-\sin \theta - (1-\cos\theta)\right]^2 \\ \\ \\ = \left[1-\sin\theta-1+\cos\theta\right]^2 \\ \\ \\ = \left[\cos \theta-\sin\theta\right]^2$


$= \cos^2\theta -2\cos \theta \sin \theta + \sin^2\theta \\ \\ \\ = (\cos^2\theta+\sin^2\theta)-\sin 2\theta \\ \\ \\ = 1-\sin2\theta \\ \\ \\ \\ \implies \boxed{\left[\frac{\cos^2\theta}{1+\sin\theta}-\frac{\sin^2\theta}{1+\cos\theta}\right]^2=1-\sin 2\theta}$


Thus, the Answer is Option 2) $\bold{1-sin \, 2\theta}$