Question

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

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

Answer 1

Answer: (A), (B), (C)

The correct options are
A $(x^{2}y{)}^{\frac{2}{3}}-(x{y}^2{)}^{\frac{2}{3}}=1$
B $\sin \theta \tan \theta =y$
C $\sin \theta \cos \theta =\frac{1}{x}$

$\cot \theta +\tan \theta =x\Rightarrow \frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }=x\Rightarrow \frac{{\cos }^2\theta +{\sin }^2\theta }{\sin \theta \cos \theta }=x\Rightarrow \sin \theta \cos \theta =\frac{1}{x}$
$sec\theta -\cos \theta =y\Rightarrow \frac{1}{\cos \theta }-\cos \theta =y\Rightarrow \frac{1-{\cos }^2\theta }{\cos \theta }=y\Rightarrow \frac{{\sin }^2\theta }{\cos \theta }=y\Rightarrow \sin \theta \tan \theta =y$
Now, ${\left(x^{2}y\right)}^{\frac{2}{3}}-{\left(x{y}^2\right)}^{\frac{2}{3}}={\left({\left(\frac{1}{\sin \theta \cos \theta }\right)}^2\frac{{\sin }^2\theta }{\cos \theta }\right)}^{\frac{2}{3}}-{\left(\frac{1}{\sin \theta \cos \theta }{\left(\frac{{\sin }^2\theta }{\cos \theta }\right)}^2\right)}^{\frac{2}{3}}={\left({\sec }^3\theta \right)}^{\frac{2}{3}}-{\left({\tan }^3\theta \right)}^{\frac{2}{3}}={\sec }^2\theta -{\tan }^2\theta =1$

Also, ${\left(x^{2}y\right)}^{\frac{1}{3}}-{\left(x{y}^2\right)}^{\frac{1}{3}}={\left({\left(\frac{1}{\sin \theta \cos \theta }\right)}^2\frac{{\sin }^2\theta }{\cos \theta }\right)}^{\frac{1}{3}}-{\left(\frac{1}{\sin \theta \cos \theta }{\left(\frac{{\sin }^2\theta }{\cos \theta }\right)}^2\right)}^{\frac{1}{3}}={\left({\sec }^3\theta \right)}^{\frac{1}{3}}-{\left({\tan }^3\theta \right)}^{\frac{1}{3}}=\sec \theta -\tan \theta \ne 1$