Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Roots of equation $x^2+px-q=0$ are $\tan 30^{\large\circ}$ and $\tan 15^{\large\circ}$,then value of $2-q-p$ is :

$\begin{array}{1 1}(A)\;0&(B)\;1\\(C)\;2&(D)\;3\end{array} $

Can you answer this question?

1 Answer

0 votes
$-p=\tan 30^{\large\circ}+\tan 15^{\large\circ}$
$-q=\tan 30^{\large\circ}.\tan 15^{\large\circ}$
$\tan 45^{\large\circ}=\tan(30+15)^{\large\circ}=\large\frac{\tan 30^{\large\circ}+\tan 15^{\large\circ}}{1-\tan 30^{\large\circ}\tan 15^{\large\circ}}$
$\Rightarrow 1=-\large\frac{p}{1+q}$$\Rightarrow -p-q=1$
$\Rightarrow 2-p-q=2+1=3$
Hence (D) is the correct answer.
answered Apr 15, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App