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sin(a+b)=3/5 cos(a-b)=12/13. find tan2a.

sin(a+b)=3/5
cos(a-b)=12/13.
find tan2a.

Grade:12

3 Answers

Aman Bansal
592 Points
11 years ago

Dear Piyush,

Cos(A+B)=4/5= 0.8
A+B =36.87 

Sin(A-B)=5/13 = 0.3846
A-B = 22.62
by adding
2A =36.87 + 22.62 = 59.49
tan 2A = tan 59.49= 1.6969

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TANAYRAJ SINGH CHOUHAN
65 Points
11 years ago
sin(a+b) = 3/5 i.e. a+b = 37 & cos(a-b) = 12/13 i.e. a-b = 22.5 These are standard values of some trigonometric functions which are used as shortcuts So , from above equations we get a=30(approx) So tan2a = tan60 = v3 If you are satisfied by my answer please click yes below the answer If not satisfied post your reply back so that i can explain better & BEST OF LUCK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
sheik arbaz
38 Points
11 years ago

Dear friend,

sin(a+b)=3/5 cos(a-b)=12/13

sin,cos are possitive .so,they lies in Q1

 

 

 

(by pythagarus rule adj side is 4)

3 5

 

4 (since sin(a+B)=3/4)

 

 

=a+b

 

therefore tan(a+b)=3/4

 

cos(a-b)=12/13

 

 

5 13 (by pythagarus rule opp'' side is 5)

 

 

12 (since cos(a-b)=12/13)

=a-b

(therefore tan(a-b)=5/12)

tan(a+b+a-b)=tan2a

=tan(a+b)+tan(a-b)

1-tan(a+b).tan(a-b)

 

= 3/4 +5/12

1-3/4 *5/12

(by solving this you get tan2a=56/33)

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