Quantitative Aptitude Ques 1120
Question: If $ cosec39{}^\circ =x, $ then the value of $ \frac{1}{cose{c^{2}}51{}^\circ }+{{\sin }^{2}}39{}^\circ +{{\tan }^{2}}51{}^\circ -\frac{1}{{{\sin }^{2}}51{}^\circ {{\sec }^{2}}39{}^\circ } $ is
Options:
A) $ \sqrt{x^{2}-1} $
B) $ \sqrt{1-x^{2}} $
C) $ x^{2}-1 $
D) $ 1-x^{2} $
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Answer:
Correct Answer: C
Solution:
- Given, $ cosec39{}^\circ =x $
Then, $ \frac{1}{cose{c^{2}}51{}^\circ }+{{\sin }^{2}}39{}^\circ +{{\tan }^{2}}51{}^\circ $
$ -\frac{1}{{{\sin }^{2}}51{}^\circ \cdot {{\sec }^{2}}39{}^\circ } $
$ ={{\sin }^{2}}51{}^\circ +{{\sin }^{2}}39{}^\circ +{{\tan }^{2}}(90{}^\circ -39{}^\circ ) $
$ -\frac{1}{{{\sin }^{2}}(90{}^\circ -39{}^\circ )\cdot {{\sec }^{2}}39{}^\circ } $
$ ={{\cos }^{2}}39{}^\circ +{{\sin }^{2}}39{}^\circ +{{\cot }^{2}}39{}^\circ -\frac{1}{{{\cos }^{2}}39{}^\circ \cdot {{\sec }^{2}}39{}^\circ } $ $ =1+{{\cot }^{2}}39{}^\circ -1 $
$ =cose{c^{2}}39{}^\circ -1 $ $ [\because 1+{{\cot }^{2}}\theta =cose{c^{2}}\theta ] $
$ =x^{2}-1 $
BETA
