Answer:

The value of tan(2A) is \frac{3}{4}[/tex]

Step-by-step explanation:

Given sin(A+B)=\frac{24}{25}sin(A+B)=

25

24

cos(A-B)=\frac{4}{5}cos(A−B)=

5

4

we have to find tan2A

If sin(A+B)=\frac{24}{25}sin(A+B)=

25

24

then PERPENDICULAR is 24 and HYPOTENUSE is 25 then

By PYTHAGORAS Theorem third side can be calculated as

25^{2}=24^{2}+Base^225

2

=24

2

+Base

2

⇒ Base is 7

Hence, tan(A+B)=\frac{24}{7}tan(A+B)=

7

24

Similarly,

if cos(A-B)=\frac{4}{5}cos(A−B)=

5

4

then base is 4 and hypotenuse is 5 then

By Pythagoras Theorem third side can be calculated as

5^{2}=4^{2}+Perpendicular^25

2

=4

2

+Perpendicular

2

⇒ Perpendicular is 3

Hence, tan(A-B)=\frac{3}{4}tan(A−B)=

4

3

∵ B>A, then A-B is negative.hence, taken negative

tan(A-B)=\frac{-3}{4}tan(A−B)=

4

−3

tan(2A)=tan((A+B)+(A-B))= \frac{tan(A+B)+tan(A-B)}{1-tan(A+B)\times tan(A-B)}tan(2A)=tan((A+B)+(A−B))=

1−tan(A+B)×tan(A−B)

tan(A+B)+tan(A−B)

⇒ tan(2A)=\frac{\frac{24}{7}+(\frac{-3}{4})}{1-\frac{24}{7}\times (\frac{-3}{4})}tan(2A)=

1−

7

24

×(

4

−3

)

7

24

+(

4

−3

)

=\frac{\frac{75}{28}}{\frac{100}{28}}=\frac{3}{4}

28

100

28

75

=

4

3