Test Index

UPSC NDA Math Model Paper 3

© examsnet.com

Question : 4 of 120

Marks: +1, -0

If

\sec \theta, \tan \theta

are the roots of the equation

a x^{2}+b x+c=0

(\sec \theta - \tan \theta) \sec \theta \tan \theta

is?

b/c
-c/b
c/b
a/b

Solution:

Given:

\sec \theta, \tan \theta

are the roots of the equation

a x^{2}+b x+c=0

Now,

Sum of roots

= \sec \theta + \tan \theta = -\frac{b}{a}

Product of roots

= \sec \theta \times \tan \theta = \frac{c}{a}

As we know that,

\sec^{2} \theta - \tan^{2} \theta = 1

\Rightarrow (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1

\Rightarrow \left(-\frac{b}{a}\right) \times (\sec \theta - \tan \theta) = 1

\therefore \sec \theta - \tan \theta = -\frac{a}{b}

Now,

(\sec \theta - \tan \theta) \sec \theta \tan \theta

= -\frac{a}{b} \times \frac{c}{a} = -\frac{c}{b}

© examsnet.com

Go to Question:

Prev Question

Next Question