Question

If $\cot \theta +\tan \theta =x$ and $\sec \theta -\cos \theta =y$, then

• A. $x\sin \theta .\cos \theta =1$
• B. ${\sin }^2\theta =y\cos \theta$
• C. $(x^{2}y{)}^{1/3}+(x{y}^2{)}^{1/3}=1$
• D. $(x^{2}y{)}^{2/3}-(x{y}^2{)}^{2/3}=1$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $x\sin \theta .\cos \theta =1$
B ${\sin }^2\theta =y\cos \theta$
D $(x^{2}y{)}^{2/3}-(x{y}^2{)}^{2/3}=1$

Given $\cot \theta +\tan \theta =x$
$\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }=x$

$\Rightarrow x=\frac{1}{\sin \theta \cos \theta }$
Also given $\sec \theta -\cos \theta =y$
$\frac{1}{\cos \theta }-\cos \theta =y$

$y=\frac{{\sin }^2\theta }{\cos \theta }$
Now, $x^{2}y=\frac{1}{{\cos }^3\theta }={\sec }^3\theta$

${xy}^2={\tan }^3\theta$

So, $(x^{2}y{)}^{2/3}-(x{y}^2{)}^{2/3}$$={\sec }^2\theta -{\tan }^2\theta =1$