Question

If sin (A + B) = 1 and tan (A – B) = 1/√3, find the value of:

1. tan A + cot B

2. sec A – cosec B​

Answer 1

Answer:

1)2root3(i didnt get root symbol in my keyboard)

2)0

Step-by-step explanation:

see its simple

sin(A+B)=1

Actually sin90 is 1

soo A+B =90----1ST EQUATION

Then it is given that

tan(A-B)=1/root3

soo tan 30 is 1/root3

therefore A-B=30-------2nd equation

A+B=90

A-B=30(B gets cancelled)

-------------

2A=120

A=120/2

THEREFORE A=60

As A+B =90

60+B=90

therefore B =30

now we found A and B

now lets do 1st subquestion

1)tanA+cotB

A we found an 60 and B we found as 30 soo substitute them

tan60 +cot 30

root3 + root 3

therefore answer is 2root3(so sorry i did not find root symbol in my keyboard)

now 2nd one

2)secA- cosecB

sec60-cosec30

2-2

=0

(u may be knowing but if u dont knowthere is a fixed values for sin ,cos,tan,cosec,cot and sec u have to byheart them)

thanks a lot for this question hope u understood

My name is Nishanya i m 15

byee

Answer 2

Answer :

1). tanA + cotB = 2√3

2). secA - cosecB = 0

Solution :

• Given : sin(A+B) = 1 , tan(A-B) = 1/√3
• To find : tanA + cotB , secA - cosecB

We have ;

=> sin(A + B) = 1

=> sin(A + B) = sin90°

=> A + B = 90° --------(1)

Also ,

=> tan(A - B) = 1/√3

=> tan(A - B) = tan30°

=> A - B = 30° ---------(2)

Now ,

Adding eq-(1) and (2) , we get ;

=> A + B + A - B = 90° + 30°

=> 2A = 120°

=> A = 120°/2

=> A = 60°

Now ,

Putting A = 60° in eq-(1) , we get ;

=> A + B = 90°

=> 60° + B = 90°

=> B = 90° - 60°

=> B = 30°

Now ,

→ tanA + cotB = tan60° + cot30°

→ tanA + cotB = √3 + √3

→ tanA + cotB = 2√3

Also ,

→ secA - cosecB = sec60° - cosec30°

→ secA - cosecB = 2 - 2

→ secA - cosecB = 0

Hence ,

• tanA + cotB = 2√3

• secA - cosecB = 0