Question

set $\frac{sin (A-B)}{sin (A+B)}=\frac{5}{7}$find the value of tan A×cot B

Answer 1

Answer :-

6

Given :-

$\cfrac{sin(A - B)}{sin(A + B)} = \dfrac{5}{7}$

To find :-

$tanA \times cotB$

SOLUTION :-

Firstly lets expand the L.H.S by the formulae

• sin(A-B) = sinA cosB + sinB cosA
• sin(A + B) = sinA cosB + sinB cosA

Substituting the values,

$\dfrac{sinA \: cosB \: - sinB \: cos \: A}{sinA \: cosB \: + sinB \: cosA \: } = \dfrac{5}{7}$

Do the cross multiplication

$(sinAcosB - sinB \: cosA)7 = 5(sinAcosB + sinB \: cosA)$

$7sinAcosB - 7sinB \: cosA = 5sinAcosB + 5sinB \: cosA$

Transposing all terms to L.H.S

$7sinAcosB - 7sinB \: cosA - 5sinAcosB - 5sinB \: cosA= 0$

Keep like terms together

$7sinAcosB -5sinAcosB -7sinBcosA- 5sinBcosA$

$2sinAcosB \: - 12sinB \: cos \: A = 0$

$2sinA \: cosB \: = \: 12sinB \: cosA$

$sinA \: cosB \: = 6sinB \: cos \: A$

$\cfrac{sinA \: cosB}{sinB \: cosA} = 6$

$\cfrac{sinA \: cosB \: }{cosA \: sinB \: } = 6$

$\cfrac{sinA}{cosA} \times \cfrac{cosB}{sinB} = 6$

From trigonometric relations,

• sinA/cosA = tanA
• cosA/sinA = cotA

$tan \: A\: cot \: B = 6$

So, the Required answer is :-

$\red{tan \: A\: cot \: B = 6}$

Know more :-

Compound angles:-

$cos(A+B) = cosAcosB - sinAsinB$

$cos(A-B) = cosAcosB + sinAsinB$

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

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

$cot(A+B) = \dfrac{cotB cotA -1}{cotB + cotA}$

$cot(A-B) = \dfrac{cotB cotA + 1}{cotB-cotA}$