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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