Quantitative Aptitude Ques 1415
Question: If $ a\sec \theta =x $ and $ b\tan \theta =y, $ then how area: and y connected with a and b?
Options:
A) $ a^{2}x^{2}-b^{2}y^{2}=a^{2}b^{2} $
B) $ b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2} $
C) $ a^{2}x^{2}+b^{2}y^{2}=a^{2}b^{2} $
D) $ b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
- Given, $ a\sec \theta =x $ and $ b\tan \theta =y $ By Hit and Trial, $ b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2} $ … (i) Now, putting the values of x and y in Eq. (i), we get $ b^{2}{{(a\sec \theta )}^{2}}-a^{2}{{(b\tan \theta )}^{2}}=a^{2}b^{2} $
$ \Rightarrow $ $ b^{2}a^{2}{{\sec }^{2}}\theta -a^{2}b^{2}{{\tan }^{2}}\theta =a^{2}b^{2} $
$ \Rightarrow $ $ a^{2}b^{2}({{\sec }^{2}}\theta -{{\tan }^{2}}\theta )=a^{2}b^{2} $
$ \therefore $ $ a^{2}b^{2}=a^{2}b^{2} $ $ [{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1] $
BETA
