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If sin (A + B) = 1 and tan (A - B) = 1/v3, find the value of:
i) tan A+ tan B
ii) sec A- cosec B
​

Respuesta :

lucky75

Given :-

If  sin (A + B) = 1

and tan (A - B) = 1/√3

To Find :-

value of:

  • i) tan A+ tan B
  • ii) sec A- cosec B

Solution :-

We know that,

  • sin 90 = 1
  • tan 30 = 1/√3

So,

  • sin(A + B) = sin90
  • (A+B)=90

  • tan(A - B) =tan 30
  • (A-B)=30

A + B + A - B = 90 + 30

(A + A) + (B - B) = 120

2A = 120

A = 120/2

  • A = 60

By putting value of A in 2

A - B = 30

60 - B = 30

-B = 30 - 60

-B = -30

  • B = 30

Finding values

tan 60 + tan 30

We know that

  • tan 60 = √3
  • tan 30 = 1/√3

√3 + 1/√3

√3 × √3 + 1/√3

3 + 1/√3

[tex]4/√3[/tex]

sec A- cosec B

sec 60 - cosec 30

We know that

  • sec 60 = 2
  • cosec 30 = 2

2 - 2 = [tex]0[/tex]

Hence,

tan A+ tan B=4/√3

sec A- cosec B=0

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