Question : $\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Option 1: $1-\cos \theta$
Option 2: $1-\sin \theta$
Option 3: $\cos \theta$
Option 4: $\sin \theta$
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Correct Answer: $1-\sin \theta$
Solution :
$\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}$
Since $ 0^{\circ}<\theta<90^{\circ}$.
$=\frac{(1+\frac{1}{\cos\theta.\sin\theta})^2(\frac{1}{cos\theta}-\frac{\sin\theta}{\cos\theta})^2 (1+\sin\theta)}{(\sin\theta+\frac{1}{\cos\theta})^2+(\cos\theta+\frac{1}{\sin\theta})^2}$
$=\frac{(\frac{\cos\theta.\sin\theta+1}{\cos\theta.\sin\theta})^2(\frac{1-\sin\theta}{\cos\theta})^2 (1+\sin\theta)}{(\frac{\sin\theta.\cos\theta+1}{\cos\theta})^2+(\frac{\cos\theta\sin\theta+1}{\sin\theta})^2}$
$=\frac{(\frac{1}{\cos\theta.\sin\theta})^2(\frac{1-\sin\theta}{\cos\theta})^2 (1+\sin\theta)}{\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta.\cos^2\theta}}$
$=\frac{(1-\sin\theta)(1-\sin\theta)(1+\sin\theta)}{\cos^2\theta}$
$=\frac{(1-\sin\theta)(1-\sin^2\theta)}{\cos^2\theta}$
$=\frac{(1-\sin\theta)(\cos^2\theta)}{\cos^2\theta}$
$=1-\sin\theta$
Hence, the correct answer is $1-\sin \theta$.
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