If x,y are acute angles, where 0 < x + y < 90° and sin(3x - 40°) = cos(3y + 40°), then the value of tan(x + y) is
equal to:
Answers
Step-by-step explanation:
Given:-
x,y are acute angles, where 0 < x + y < 90° and sin(3x - 40°) = cos(3y + 40°)
To find:-
Find the value of tan(x + y) ?
Solution:-
Method-1:-
Given that :-
x,y are acute angles, where 0 < x + y < 90° and
sin(3x - 40°) = cos(3y + 40°)
It can be written as
=>Sin [90°-(-3x+130°)] = Cos (3y+40°)
=>We know that Sin (90°-A)=Cos A
=>Cos (-3x+130°) = Cos (3y+40°)
Since x and y are acute angles
=>-3x +130° = 3y +40°
=>3x+3y = 130°-40°
=>3x+3y = 90°
=>3(x+y)=90°
=>x+y = 90°/3
=>x+y = 30°
Now Tan (x+y) = Tan 30°
Tan(x+y)=1/√3
Method-2:-
Given that :-
x,y are acute angles, where 0 < x + y < 90° and
sin(3x - 40°) = cos(3y + 40°)
It can be written as
Sin (3x-40°) = Cos [90°-(-3y+50°)]
We know that
Cos (90°-A)=SinA
=>Sin (3x-40°) = Sin(-3y+50°)
Since x and y are acute angles
=>3x-40° =-3y +50°
=>3x+3y = 50°+40°
=>3x+3y = 90°
=>3(x+y)=90°
=>x+y = 90°/3
=>x+y = 30°
On taking tan both sides then
=>Tan(x+y)=Tan 30°
=>Tan(x+y)=1/√3
Answer:-
The value of Tan(x+y) for the given problem is
1/√3
Check:-
If x+y=30° then x = 30°-y LHS
Sin[ (3×30°-y)-40°)]
=>Sin (90°-3y-40°)
=>Sin (90°-(3y+40°)
=>Cos (3y+40°)
=>RHS
Verified the given relation
Used formulae:-
- Sin (90°-A)=Cos A
- Cos (90°-A)=SinA
- Tan30°=1/√3
Answer:
refer the attachment please hope it helps
