Question

If sin(a+b)=1 and tan(a-b)= 1/√3 find tan a + cot b and sec a - cosec b

Answer 1

Answer:

tanA + CotB = 2√3 and SecA - CosecB = 0

Step-by-step explanation:

Given---> Sin( A + B ) = 1 and tan ( A - B ) = 1 /√3

To find---> tanA + CotB = ? and

SecA - CosecB = ?

Solution---> ATQ,

Sin ( A + B ) = 1

=> Sin( A + B ) = Sin 90°

=> A + B = 90° ..........................(1)

tan ( A - B ) = 1 / √3

=> tan ( A - B ) = tan 30°

=> A - B = 30°

=> A = 30° + B ......................(2)

Putting A = 30° + B in equation ( 1 ) , we get

=> 30° + B + B = 90°

=> 2B = 90° - 30°

=> 2B = 60°

=> B = 60° / 2

=> B = 30°

Putting B = 30° in equation (2) , we get

=> A = 30° + B

=> A = 30° + 30°

=> A = 60°

Now , tanA + CotB = tan60° + Cot30°

= √3 + √3

= 2 √3

Now, SecA - CosecB = Sec 60° - Cosec30°

= 2 - 2

= 0