NIELIT 2019 Feb Scientist C - Section D: 16

Question

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

• A. $1$
• B. $2$
• C. $2 \sin\theta$
• D. $\sin\theta \cos\theta$

Answer 1

Given that:

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

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

 substitute $l$ & $m$ value in given equation;

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

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

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

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

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

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

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

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

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

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

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

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

$\implies 1$

Option (a) is correct.