2sinAcosA+(cosA+sin)2-(2cosA+sinA)2 = psin2A +q .find the value of p and q
Answers
Step-by-step explanation:
Given:-
2SinACosA+(CosA+SinA)²-(2CosA+SinA)² = p Sin 2A + q
To find :-
Find the value of p and q ?
Solution :-
Given that
2 Sin A Cos A+(Cos A + Sin A)²-(2 Cos A + Sin A)² = p Sin 2A + q
On taking LHS
2SinACosA+(CosA+SinA)²-(2CosA+ SinA)²
We know that
Sin 2A = 2 Sin A Cos A ------(1)
and
(Cos A + Sin A)²
= Cos² A + 2 Sin A Cos A + Sin² A
Since , (a+b)² = a²+2ab+b²
=> (Sin² A + Cos² A ) + 2 Sin A Cos A
=> 1 + 2 Sin A Cos A
=> 1 + Sin 2 A ---------(2)
and
(2 Cos A + Sin A)²
=> 4 Cos² A + 4 Cos A Sin A +Sin² A
Since, (a+b)² = a²+2ab+b²
=>Cos²A+3Cos²A+2(2SinACosA)+Sin²A
=>(Sin²A+Cos²A)+2(2SinACosA)+3Cos²A
=> 1+ 2 Sin 2A + 3 Cos² A ------(3)
On adding (1)&(2)
Sin 2A + 1 + Sin 2 A
=> 2 Sin 2A + 1 ------------(4)
On subtracting (3) from (4)
=> (2 Sin 2A + 1 ) - (1+ 2 Sin 2A + 3 Cos² A)
=> 2 Sin 2A + 1 - 1 - 2 Sin 2A - 3 Cos² A
=> -3 Cos² A
=> -3 ( 1 - Sin² A)
=> -3 + 3 Sin² A
=> 3 Sin² A - 3
=> 3 Sin² A +(-3)
Now
We have ,
LHS = 3 Sin² A +(-3)
2SinACosA+(CosA+SinA)²-(2CosA+Sin A)²
= 3 Sin² A +(-3)
Now
We have,
3 Sin² A + (-3) = p Sin² A + q
On comparing both sides then
p = 3 and q = -3
Answer:-
The values of p and q are 3 and -3 respectively.
Used formulae:-
→ Sin 2A = 2 Sin A Cos A
→ Sin² A + Cos² A = 1
→ (a+b)² = a²+2ab+b²