Question

7tan(a-b)=5tan(a+b) then find the value of sin2a\sin2b​

Answer 1

$\textbf{Given:}$

$7\,tan(A-B)=5\,tan(A+B)$

$\textbf{To find:}$

$\text{The value of \dfrac{sin2A}{sin2B} }$

$\textbf{Solution:}$

$\text{Consider,}$

$7\,tan(A-B)=5\,tan(A+B)$

$\implies\,\dfrac{tan(A-B)}{tan(A+B)}=\dfrac{5}{7}$.....(1)

$\text{Now,}$

$\dfrac{sin2A}{sin2B}$

$=\dfrac{sin[(A+B)+(A-B)}{sin[(A+B)-(A-B)]}$

$=\dfrac{sin(A+B)\,cos(A-B)+cos(A+B)\,sin(A-B)}{sin(A+B)\,cos(A-B)-cos(A+B)\,sin(A-B)}$

$\text{Divide both numerator and denominator by sin(A+B)\,cos(A-B) }$

$=\dfrac{1+cot(A+B)\,tan(A-B)}{1-cot(A+B)\,tan(A-B)}$

$=\dfrac{1+\dfrac{tan(A-B)}{tan(A+B)}}{1-\dfrac{tan(A-B)}{tan(A+B)}}$

$\text{Using (1)}$

$=\dfrac{1+\dfrac{5}{7}}{1-\dfrac{5}{7}}$

$=\dfrac{\dfrac{7+5}{7}}{\dfrac{7-5}{7}}$

$=\dfrac{\dfrac{12}{7}}{\dfrac{2}{7}}$

$=\dfrac{12}{2}$

$=6$

$\therefore\textbf{The value of \bf\,\dfrac{sin2A}{sin2B} is 6}$

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Answer 2

Step-by-step explanation:

given 7tan(a-b)=5tan(a+b) then find the value of sin2a\sin2b

• Given 7 tan (a – b) = 5 tan (a + b)      
•   so tan (a – b) / tan (a + b) = 5/7
•  Now sin (a – b) / cos (a – b)  / sin (a + b) / cos (a + b) = 5/7
•         Sin (a – b) x cos(a + b) / cos (a – b) x sin(a + b) = 5/7
•     Now we can write as  
•            2 sin(a – b) cos (a + b) / 2 cos(a – b) sin (a + b) = 5/7
•          [sin 2a – sin 2b / sin 2a + sin 2b = 5/7
•        So we get  Sin 2a – sin 2b = 2 sin (a – b) cos (a + b)]
•           So  Sin 2a + sin 2b = 2 cos(a – b) sin (a + b)
•              By taking reciprocal we get
•             (sin 2a + sin 2b) / (sin 2a – sin 2b) = 7/ 5
• Now by comp 6 + 1 / 6 – 1
•             Sin 2a / sin 2b = 6 / 1
• Therefore sin 2a / sin 2b = 6

Reference link will be

https://brainly.in/question/16648416