Question

If 0 < A < B < $\frac{\pi}{4}$ and sin(A + B) = $\frac{24}{25}$ and cos(A - B) = $\frac{4}{5}$, then find the value of tan 2A.

Answer 1

given, If 0 < A < B < $\frac{\pi}{4}$ and sin(A + B) = $\frac{24}{25}$ and cos(A - B) = $\frac{4}{5}$.

sin(A + B) = 24/25

so, cos(A + B) = 7/25

again, cos(A - B) = 4/5

so, sin(A - B) = 3/5


now, sin2A = sin{(A + B) + (A - B)}

= sin(A + B).cos(A - B) + sin(A - B) cos(A + B)

= 24/25 × 4/5 + 7/25 × 3/5

= 96/125 + 21/125

= 117/125

sin2A = 117/125 = p/h

so, p = 117 and h = 125

then, b = √(h² - p²) = √(125² - 117²)

= √(125 + 117)(125 - 117) = √(242 × 8)

= 11 × 4 = 44

hence, tan2A = p/b = 117/44

Answer 2

HELLO DEAR,




sin(A + B) = 24/25 

so, cos(A + B) = 7/25 

andcos(A - B) = 4/5 

so, sin(A - B) = 3/5 


now,
sin2A = sin{(A + B) + (A - B)} 

=> sin(A + B).cos(A - B) + sin(A - B) cos(A + B) 

=> 24/25 × 4/5 + 7/25 × 3/5 

=> 96/125 + 21/125 

=> 117/125 

sin2A = 117/125 = p/h 

so, p = 117 and h = 125 

then, b = √(h² - p²) = √(125² - 117²) 

= √(125 + 117)(125 - 117) = √(242 × 8) =√(121 × 16)

= 11 × 4 = 44 

hence, tan2A = p/b = 117/44


I HOPE IT'S HELP YOU DEAR,
THANKS