Quantitative Aptitude Ques 27
Question: The value of is $ \cot \theta \cdot \tan ,(90{}^\circ -\theta )-sec,(90{}^\circ -\theta ) $ $ cosec\theta +({{\sin }^{2}}25{}^\circ +{{\sin }^{2}}65{}^\circ )+ $ $ \sqrt{3},(\tan 5{}^\circ \cdot \tan 15{}^\circ \cdot \tan 30{}^\circ \cdot \tan 75{}^\circ \cdot \tan 85{}^\circ ) $ is
Options:
A) 1
B) $ -1 $
C) 2
D) 0
Show Answer
Answer:
Correct Answer: A
Solution:
- Expression =
$ =\cot \theta \cdot \tan ,(90{}^\circ -\theta )-sec,(90{}^\circ -\theta ) $
$ cosec\theta +({{\sin }^{2}}25{}^\circ +{{\sin }^{2}}65{}^\circ )+\sqrt{3} $
$ (\tan 5{}^\circ \cdot tan15{}^\circ \cdot \tan 30{}^\circ \cdot \tan 75{}^\circ \cdot \tan 85{}^\circ ) $
$ =\cot \theta \cdot \cot \theta -cosec\theta \cdot cosec\theta + $
$ ({{\sin }^{2}}25{}^\circ +{{\cos }^{2}}25{}^\circ )+\sqrt{3}(\tan 5{}^\circ $
$ \cot 5{}^\circ \cdot \tan 15{}^\circ \cot 15{}^\circ \cdot \tan 30{}^\circ ) $
$ \begin{bmatrix} \because \sec ,(90{}^\circ -\theta )=cosec\theta \\ \sin ,(90{}^\circ -\theta )=\cos \theta \\ \tan ,(90{}^\circ -\theta )=\cot \theta \\ \end{bmatrix} $
$ =({{\cot }^{2}}\theta -cose{c^{2}}\theta )+({{\sin }^{2}}25{}^\circ +{{\cos }^{2}}25{}^\circ )+\sqrt{3}\times \frac{1}{\sqrt{3}} $ $ =-1+1+1=1 $
$ [\because {{\sin }^{2}}\theta {{\cos }^{2}}\theta =1,and,{{\cot }^{2}}\theta -cose{c^{2}}\theta =-1] $
BETA
