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

Since $\cot \theta +\tan \theta =x$ and
$\sec \theta -\cos \theta =y$
option (A):
$\therefore \frac{1}{x}=\frac{1}{\cot \theta +\tan \theta }=\frac{sin\theta cos\theta }{{\cos }^2\theta +{\sin }^2\theta }$
$=\frac{\sin \theta \cos \theta }{1}$
$\therefore \sin \theta \cos \theta =\frac{1}{x}$
option (B):
since, $y=\sec \theta -\cos \theta$
$=\frac{1-{\cos }^2\theta }{\cos \theta }=\frac{{\sin }^2\theta }{\cos \theta }$
$\sin \theta \tan \theta$
Hence, $\sin \theta \tan \theta =y$
option (C):
since, $x^{2}y$
$=\frac{1}{{\sin }^2\theta {\cos }^2\theta }\times \frac{{\sin }^2\theta }{\cos \theta }=\frac{1}{{\cos }^3\theta }$
$\Rightarrow (\frac{1}{x^{2}y}{)}^{\frac{2}{3}}={\cos }^2\theta$
$\Rightarrow (x^{2}y{)}^{\frac{2}{3}}={\sec }^2\theta ....(i)$
and $x{y}^2=\frac{1}{\sin \theta \cos \theta }\times \frac{{\sin }^4\theta }{{\cos }^2\theta }={\tan }^3\theta$
$\Rightarrow (x{y}^2{)}^{\frac{2}{3}}={\tan }^2\theta ....(ii)$
From equation (i) and (ii),
$(x^{2}y{)}^{\frac{2}{3}}-(x{y}^2{)}^{\frac{2}{3}}=1$