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If $cosec\theta-\sin\theta=1$ and $\sec\theta-\cos\theta=m$, then $l^{2}m^{2}(l^{2}+m^{2}+3)$ equals to:

  1. $1$
  2. $2$
  3. $2 \sin\theta$
  4. $\sin\theta \cos\theta$
in Quantitative Aptitude recategorized by
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$\textrm{Given that:}$

$\csc \theta-\sin\theta =l$

$\sec\theta-\cos\theta=m$

$\textrm{ substitute l & m value in given equation;}$

$\implies$ $(\csc\theta-\sin\theta)^2*(\sec\theta-\cos\theta)^2*\left((\csc\theta-\sin\theta)^2+(\sec\theta-\cos\theta)^2+3\right)$

$\because \sin\theta=\frac{1}{\csc\theta},\cos\theta=\frac{1}{\sec\theta}$

$\implies$ $(\frac{1}{\sin\theta}-\sin\theta)^2*(\frac{1}{\cos\theta}-\cos\theta)^2*\left((\frac{1}{\sin\theta}-\sin\theta)^2+(\frac{1}{\cos\theta}-\cos\theta)^2+3\right)$

$\implies$  $(\frac{1-\sin^2\theta}{\sin\theta})^2*(\frac{1-cos^2\theta}{\cos\theta})^2*\left((\frac{1-sin^2\theta}{\sin\theta})^2+(\frac{1-cos^2\theta}{\cos\theta})^2+3\right)$

$\because sin^2\theta+cos^2\theta=1$

$\therefore$ $\implies$ $(\frac{\cos^2\theta}{\sin\theta})^2(\frac{sin^2\theta}{\cos\theta})^2\left((\frac{\cos^2\theta}{\sin\theta})^2+(\frac{sin^2\theta}{\cos\theta})^2+3\right)$

$\implies$ $\frac{\cos^4\theta}{\sin^2\theta}*\frac{sin^4\theta}{\cos^2\theta}* \left(\frac{cos^4\theta}{\sin^2\theta}+\frac{sin^4\theta}{cos^2\theta}+3\right)$

$\implies$ $\sin^2\theta\cos^2\theta*\left(\frac{\cos^6\theta+\sin^6\theta+3\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta}\right)$

$\implies$ $(\cos\theta^2)^3+(\sin\theta^2)^3+3\sin^2\theta cos^2\theta$

$\because a^3+b^3=(a+b)^3-3ab(a+b)$

$\therefore$ $\implies$ $(sin^2\theta+cos^2\theta)^3-3sin^2\theta cos^2\theta(sin^2\theta+cos^2\theta)+3sin^2\theta cos^2\theta$

$\implies$ $1-3sin^2\theta cos^2\theta+3sin^2\theta cos^2\theta$

$\implies 1$

$\textrm{option a is correct.}$
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