NTSE Stage 2 SAT 2015 Question Paper

$\mathrm{cosec}\mathrm{x}-\cot \mathrm{x}=\frac{1}{3}$
∴ $\frac{1}{\sin \mathrm{x}}-\frac{\cos \mathrm{x}}{\sin \mathit{x}}=\frac{1}{3}$
∴ $\frac{1-\cos \mathrm{x}}{\sin \mathrm{x}}=\frac{1}{3}$
∴ $3-3\cos \mathrm{x}=\sin \mathrm{x}$
Now squaring on both side, we get
$9+9{\cos }^2\mathrm{x}-18\cos \mathrm{x}={\sin }^2\mathrm{x}$
$9{\cos }^2\mathrm{x}-18\cos \mathrm{x}+9=1-{\cos }^2\mathrm{x}$
$\therefore 10{\cos }^2\mathrm{x}-18\cos \mathrm{x}+8=0$
∴ $5{\cos }^2\mathrm{x}-9\cos \mathrm{x}+4=0$
$\therefore 5{\cos }^2\mathrm{x}-5\cos \mathrm{x}-4\cos \mathrm{x}+4=0$
$\therefore \left(5\cos \mathrm{x}-4\right)\left(\cos \mathrm{x}-1\right)=0$
∴ $\cos \mathrm{x}=\frac{4}{5}\mathrm{or}\cos \mathrm{x}=1$
Now for $\cos \mathrm{x}=\frac{4}{5},{\sin }^2\mathrm{x}=1-{\cos }^2\mathrm{x}$
$=1-\frac{16}{25}=\frac{9}{25}$
$\therefore {\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$ and for $\cos \mathrm{x}=1$(Which is not possible as $\cos \mathrm{x}-1$ for x=0 )
Hence option (d) is correct.