Trigonometric Identities Question 178
Question: If $ \cot ,\theta +\tan \theta =m $ and $ \sec \theta -\cos \theta =n, $ then which of the following is correct
Options:
A) $ m{{(mn^{2})}^{1/3}}-n{{(nm^{2})}^{1/3}}=1 $
B) $ m{{(m^{2}n)}^{1/3}}-n{{(mn^{2})}^{1/3}}=1 $
C) $ n{{(mn^{2})}^{1/3}}-m{{(nm^{2})}^{1/3}}=1 $
D) $ n{{(m^{2}n)}^{1/3}}-m{{(mn^{2})}^{1/3}}=1 $
Show Answer
Answer:
Correct Answer: A
Solution:
As given $ \frac{1}{\tan \theta }+\tan \theta =m,\Rightarrow ,1+{{\tan }^{2}}\theta =m,\tan \theta $
$ \Rightarrow {{\sec }^{2}}\theta =m,\tan \theta $ ?..(i) and $ \sec \theta -\cos \theta =n\Rightarrow {{\sec }^{2}}\theta -1=n,\sec \theta $
$ \Rightarrow {{\tan }^{2}}\theta =n\sec \theta $
$ \Rightarrow {{\tan }^{4}}\theta =n^{2},{{\sec }^{2}}\theta =n^{2}.,m\tan \theta $ {by (i)}
$ \Rightarrow {{\tan }^{3}}\theta =n^{2}m,,,(,\because \tan \theta \ne 0) $
$ \Rightarrow \tan \theta ={{(n^{2}m)}^{1/3}} $ ?..(ii) Also, $ {{\sec }^{2}}\theta =m\tan \theta =m,{{(n^{2}m)}^{1/3}} $ {by (i) and (ii)} \ Using the identity $ {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 $
$ \Rightarrow m,{{(mn^{2})}^{1/3}}-{{(n^{2}m)}^{2/3}}=1 $
$ \Rightarrow m,{{(mn^{2})}^{1/3}}-n,{{(nm^{2})}^{1/3}}=1. $
BETA
