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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
Answers
EXPLANATION.
⇒ sin(A + B) = 1. - - - - - (1).
⇒ tan(A - B) = 1/√3. - - - - - (2).
As we know that,
We can write equation as,
⇒ sin(A + B) = sin(90°). - - - - - (1).
⇒ tan(A - B) = tan(30°). - - - - - (2).
We get,
⇒ A + B = 90°. - - - - - (1).
⇒ A - B = 30°. - - - - - (2).
We get,
⇒ 2A = 120°.
⇒ A = 60°.
Put the value of A = 60° in equation (1), we get.
⇒ 60° + B = 90°.
⇒ B = 90° - 60°.
⇒ B = 30°.
Value of A = 60° & B = 30°.
To find :
(1) = tan(A) + tan(B).
⇒ tan(60°) + tan(30°).
⇒ √3 + 1/√3.
⇒ 3 + 1/√3 = 4/√3.
tan(A) + tan(B) = 4/√3.
(2) = sec(A) - cosec(B).
⇒ sec(60°) - cosec(30°).
⇒ 2 - 2 = 0.
sec(A) - cosec(B) = 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