If seca and coseca are the roots of
x2+px+q=0
(A) P2=q(q-2)
(B) P2=q(q+2)
(C) p2+q2=2q
(D) None
Answers
Step-by-step explanation:
Given:-
sec a and cosec a are the roots of x²+px+q=0
To prove:-
If sec a and cosec a are the roots of x²+px+q=0
Which of the following is correct:
(A) P²=q(q-2)
(B) P²=q(q+2)
(C) p²+q²=2q
(D) None
Solution:-
Given quadratic equation is x²+px+q=0
On comparing with the standard quadratic equation ax²+bx+c=0,
we have,
a=1;
b=p;
c=q
Given roots of the quadratic equation are 'Sec a' and 'Cosec a'
Let α=Sec a
and Let β= Cosec a
we know that,
Sum of the roots =-b/a
α+β=-b/a
=>Sec a+Cosec a= -p/1
=>Sec a+Cosec a= -p
=>(1/Cos a)+(1/Sin a)= -p
=>(Cos a+Sin a)/(Cos a Sin a)= -p
=>(Sin a+Cos a)/Sina Cosa= -p----------------(1)
Product of the roots=c/a
αβ=q/1
Sec a×Cosec a=q
(1/Cos a)×(1/Sin a)=q
=>1/Sin a Cos a=q------------------(2)
and Sin a Cos a=1/q --------------(3)
Substituting the value of "q" in equation(1) then
=>(Sin a+Cos a)(q)= -p
=>Sin a+ Cos a=-p/q
On squaring both sides ,then
=>(Sin a+Cos a)²=(-p/q)²
(since (a+b)²=a²+2ab+b²)
=>Sin²a+Cos²a+2 sinacosa=p²/q²
(since sin² a+cos² a=1)
=>1+2 sin a cos a=p²/q²
=>1+2(1/q)=p²/q²
=>1+(2/q)=p²/q²
=>(q+2)/q=p²/q²
On applying cross multiplication, then
=>p²×q=q²×(q+2)
(since q is cancelled on both sides)
therefore,p²=q(q+2)
Answer:-
Option B is the correct answer .i.e. p²=q(q+2)
Using formulae:-
- Let α and β are the roots of the quadratic equation ax²+bx+c=0 then,
- Sum of the roots =α+β=-b/a
- Product of the roots=αβ=c/a
- Sec a= 1/Cos a
- Cosec a=1/Sin a
- (a+b)²=a²+2ab+b²
- Sin² a+Cos²a=1