Please Answer my attachment question..
Answers
Answer:
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) = 90
tan(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
4/√3
sec A- cosec B
sec 60 - cosec 30
We know that
sec 60 = 2
cosec 30 = 2
2 - 2 = 0
..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) = 90
tan(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
4/√3
sec A- cosec B
sec 60 - cosec 30
We know that
sec 60 = 2
cosec 30 = 2
2 - 2 = 0
..
_________✅ᴀɴsᴡᴇʀ ✅________
Sin c= P/H
perpendicular = √(13)^2-(5)^2
= √144= 12
= 12/13
Cosc = √1-sin^2c
1-144/169= 5/13
TanC= sinc/cosc = 12/13/(5/13)
= 12/5