Question : If $\operatorname{cosec}\theta-\sin\theta=l$ and $\sec\theta-\cos\theta=m$, then the value of $l^2m^2(l^2+m^2+3)$ is:
Option 1: $–1$
Option 2: $0$
Option 3: $1$
Option 4: $2$
Correct Answer: $1$
Solution : Given: $\operatorname{cosec}\theta-\sin\theta=l$ and $\sec\theta-\cos\theta=m$, $l^2m^2(l^2+m^2+3)$ $=(\operatorname{cosec}\theta-\sin\theta)^2(\sec\theta-\cos\theta)^2[(\operatorname{cosec}\theta-\sin\theta)^2+(\sec\theta-\cos\theta)^2+3]$ $= (\frac{1}{\sin\theta}-\sin\theta)^2(\frac{1}{\cos\theta}-\cos\theta)^2[(\frac{1}{\sin\theta}-\sin\theta)^2+(\frac{1}{\cos\theta}-\cos\theta)^2+3]$ $=(\frac{1-\sin^2\theta}{\sin\theta})^2(\frac{1-\cos^2\theta}{\cos\theta})^2[(\frac{1-\sin^2\theta}{\sin\theta})^2+(\frac{1-\cos^2\theta}{\cos\theta})^2+3]$ $=(\frac{cos^2\theta}{\sin\theta})^2(\frac{\sin^2\theta}{\cos\theta})^2[(\frac{\cos^2\theta}{\sin\theta})^2+(\frac{\sin^2\theta}{\cos\theta})^2+3]$ $=(\frac{\cos^4\theta}{\sin^2\theta})(\frac{\sin^4\theta}{\cos^2\theta})[(\frac{\cos^4\theta}{\sin^2\theta})+(\frac{\sin^4\theta}{\cos^2\theta})+3]$ $= \sin^2\theta \cos^2\theta[\frac{\cos^6\theta+\sin^6\theta+3\sin^2\theta \cos^2\theta}{\sin^2\theta \cos^2\theta}]$ $= \cos^6\theta+\sin^6\theta+3\sin^2\theta \cos^2\theta$ $=(\cos^2\theta)^3+\sin^2\theta)^3+3\sin^2\theta \cos^2\theta$ $=(\cos^2\theta+\sin^2\theta)^3-3\cos^2\theta \sin^2\theta(\cos^2\theta+\sin^2\theta)+3\sin^2\theta \cos^2\theta$ $=1^3-3\cos^2\theta \sin^2\theta(1)+3\sin^2\theta \cos^2\theta$ $=1$ Hence, the correct answer is $1$.
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Question : If $\operatorname{cos} \theta+\operatorname{sin} \theta=\sqrt{2} \operatorname{cos} \theta$, find the value of $(\cos \theta-\operatorname{sin} \theta)$
Option 1: $\sqrt{2} \sin \theta$
Option 2: $\sqrt{2} \cos \theta$
Option 3: $\frac{1}{\sqrt{2}} \sin \theta$
Option 4: $\frac{1}{2}\cos \theta$
Question : If $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ is:
Option 1: 0
Option 2: 2
Option 3: 5
Option 4: 1
Question : If $\frac{1}{\operatorname{cosec} \theta+1}+\frac{1}{\operatorname{cosec} \theta-1}=2 \sec \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\tan \theta+2 \sec \theta}{\operatorname{cosec} \theta}$ is:
Option 1: $\frac{4+\sqrt{2}}{2}$
Option 2: $\frac{2+\sqrt{3}}{2}$
Option 3: $\frac{4+\sqrt{3}}{2}$
Option 4: $\frac{2+\sqrt{2}}{2}$
Question : If $\theta$ is an acute angle and $\sin \theta+\operatorname{cosec} \theta=2$, then the value of $\sin ^5 \theta+\operatorname{cosec}^5 \theta$ is:
Option 1: 10
Option 3: 4
Option 4: 5
Question : If $\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4,0^{\circ}<\theta<90^{\circ}$, then what is the value of $(\sec \theta+\operatorname{cosec} \theta+\cot \theta) ?$
Option 1: $1+2 \sqrt{3}$
Option 2: $\frac{1+2 \sqrt{3}}{3}$
Option 3: $\frac{2+\sqrt{3}}{3}$
Option 4: $2+\sqrt{3}$
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