if a and b are acute angels and sin(a+b)=1,sin(a-b)=1/2
then prove that
(1)tan(a+2b)=-(3)1/2
To prove that tan(a + 2b) = -√3, we will use the given information and trigonometric identities.
1. Start by using the identity sin(a + b) = sin(a)cos(b) + cos(a)sin(b):
sin(a + b) = sin(a)cos(b) + cos(a)sin(b) = 1
2. Since sin(a) and sin(b) have maximum values of 1, and cos(a) and cos(b) have maximum values of 1, we infer that both sin(a) and sin(b) must be 1, and both cos(a) and cos(b) must be positive.
3. Using the second given equation, sin(a - b) = 1/2, we can rewrite it as sin(a)cos(b) - cos(a)sin(b) = 1/2.
4. Combining equations from step 1 and step 3, we have:
sin(a)cos(b) + cos(a)sin(b) = 1 ...(1)
sin(a)cos(b) - cos(a)sin(b) = 1/2 ...(2)
5. Add equations (1) and (2) to eliminate the terms with sin(b):
2sin(a)cos(b) = 3/2
6. Divide both sides of the equation by 2cos(b):
sin(a)/cos(b) = 3/4
7. Since sin(a)/cos(b) is equivalent to tan(a), we can rewrite the equation as:
tan(a) = 3/4
8. Now, we need to find the value of tan(2b). We can use the double-angle formula for tangent:
tan(2b) = 2tan(b) / (1 - tan^2(b))
9. Since we know sin(b) = 1, and cos(b) = √(1 - sin^2(b)) = √(1 - 1^2) = √0 = 0, we find that tan(b) is undefined.
10. However, if we consider the range of tan(b), which is undefined at b = π/2 and b = -π/2, and given a + b = π/2 (since sin(a+b) = 1 implies a + b = π/2), we can conclude that a must be -π/2.
11. Substituting a = -π/2 into tan(a) = 3/4, we have:
tan(-π/2) = 3/4
12. Using the periodicity of tangent, tan(-π/2) is equivalent to tan(3π/2) and by symmetry, tan(3π/2) = -√3.
13. Finally, substituting tan(3π/2) = -√3 and tan(b) = -√3 into the equation for tan(2b), we get:
tan(a + 2b) = tan(-π/2 + 2b) = tan(3π/2) = -√3
Therefore, we have proven that tan(a + 2b) = -√3, as required.