Question
If $$\sec \alpha $$ and $${\text{cosec}}\,\alpha $$ are the roots of $${x^2} - px + q = 0$$ then
A.
$${p^2} = q\left( {q - 2} \right)$$
B.
$${p^2} = q\left( {q + 2} \right)$$
C.
$${p^2} + {q^2} = 2q$$
D.
None of these
Answer :
$${p^2} = q\left( {q + 2} \right)$$
Solution :
$$\eqalign{
& \sec \alpha + {\text{cosec}}\,\alpha = p\,\,{\text{and}}\sec \alpha \cdot {\text{cosec}}\,\alpha = q. \cr
& \therefore \,\,\sin\alpha + \cos\alpha = p\sin \alpha \cdot \cos \alpha \,\,{\text{and }}\sin \alpha \cdot \cos \alpha = \frac{1}{q}. \cr
& \therefore \,\, \sin \alpha + \cos \alpha = \frac{p}{q} \cr
& \therefore \,\,\frac{{{p^2}}}{{{q^2}}} = 1 + 2\sin \alpha \cdot \cos \alpha = 1 + \frac{2}{q}. \cr} $$