. 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
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If  sin (A + B) = 1

and tan (A - B) = 1/√3

value of:

We know that,

So,

A + B + A - B = 90 + 30

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

2A = 120

A = 120/2

By putting value of A in 2

A - B = 30

60 - B = 30

-B = 30 - 60

-B = -30

tan 60 + tan 30

We know that

√3 + 1/√3

√3 × √3 + 1/√3

3 + 1/√3

sec A- cosec B

sec 60 - cosec 30

We know that

2 - 2 =

tan A+ tan B=4/√3

sec A- cosec B=0

ca = ec

cpctc makes it so that corresponding parts of the triangle you proved to be equal are equal, the other options aren't part of the part of the triangles you proved to be equal.