if (2 sin A + cosecc A)=2root2
Value of
2(sin to the power 4+cos to the power 4)
Answers
Step-by-step explanation:
Given :-
2 Sin A + Cosec A = 2√2
To find :-
Find the value of 2(Sin^4 A + Cos^4 A)?
Solution :-
Given that :
2 Sin A + Cosec A = 2√2
=> 2 Sin A + (1/Sin A) = 2√2
=> [2 Sin A( Sin A ) + 1] / Sin A = 2√2
=> (2 Sin^2 A+1)/ Sin A = 2√2
=> 2 Sin^2 A +1 = 2√2 Sin A
=> 2 Sin^2 A + 1 - 2√2 Sin A = 0
=> 2 Sin^2 A - 2√2 Sin A + 1 = 0
=> 2 Sin^2 A - √2 Sin A -√2 Sin A +1 = 0
=> √2 Sin A (√2 Sin A - 1) -1( √2 Sin A -1) = 0
=> (√2 Sin A - 1) (√2 Sin A -1) = 0
=> (√2 Sin A -1)^2 = 0
=> √2 Sin A -1 = 0
=> √2 Sin A = 1
=> Sin A = 1/√2 ---------(1)
On squaring both sides then
=> Sin^2 A = (1/√2)^2
=> Sin^2 A = 1/2
On Subtracting this from 1 both sides then
=> 1 - Sin^2 A = 1 - (1/2)
=> 1 - Sin^2 A = (2-1)/2
=> 1 - Sin^2 A = 1/2
We know that
Sin^2 A + Cos^2 A = 1
=> Cos^2 A = 1/2
=> Cos A = √ (1/2)
=> Cos A = 1/√2 ---------(2)
Now
The value of 2(Sin^4 A + Cos^4 A)
=> 2[ (1/√2)^4 + (1/√2)^4]
=> 2[(1/4)+(1/4)]
=> 2(1+1)/4
=>2(2/4)
=> 4/4
=> 1
(or)
2(Sin^4 A + Cos^4 A)
=> 2[(Sin^2 A)^2+ (Cos^2 A)^2 ]
=> 2[(Sin^2 A+ Cos^2 A)^2 - 2 Sin^2 ACos^2 A]
Since a^2+b^2 = (a+b)^2-2ab
=> 2[(1)^2-2 Sin^2 ACos^2 A]
=> 2[1-2 Sin^2 ACos^2 A]
=> 2[1-2(1/√2)^2(1/√2)^2]
=> 2[1-2(1/2)(1/2)]
=> 2[1-2(1/4)]
=> 2[1-(2/4)]
=> 2[1-(1/2)]
=> 2[(2-1)/2]
=> 2[1/2]
=> 2/2
=> 1
Answer:-
The value of 2(Sin^4 A + Cos^4 A) for the given problem is 1
Used formulae:-
- Sin^2 A + Cos^2 A = 1
- (a/b)^m = a^m /b^m
- a^2+b^2 = (a+b)^2-2ab
Rough work :-
(1/√2)^4
= 1^4/(√2)^4
= 1/(√2×√2×√2×√2)
= 1/(2×2)
= 1/4
Step-by-step explanation:
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