Problem 1 :
Prove that
(i) tan(45° + A) = (1 + tanA)/(1 − tanA)
(ii) tan(45° − A) = (1 + tanA)/(1 − tanA)
Problem 2 :
Prove :
cot(A + B) = (cotAcotB − 1)/(cotA + cotB)
Problem 3 :
If tan x = n/(n + 1) and tan y = 1/(2n + 1) , find tan(x + y).
Problem 1 :
Show that
(i) tan(45° + A) = (1 + tanA)/(1 − tanA)
(ii) tan(45° − A) = (1 + tanA)/(1 − tanA)
Solution :
(i) tan(45° + A) = (1 + tanA)/(1 − tanA)
tan (A + B) = (tan A + tan B) / (1 - tan A tan B)
= (tan 45° + tan A) / (1 - tan 45° tan A)
The value of tan 45° is 1.
= (1 + tan A) / (1 - tan A)
(ii) tan(45° − A) = (1 + tanA)/(1 − tanA)
Solution :
tan (A - B) = (tan A - tan B) / (1 + tan A tan B)
= (tan 45° - tan A) / (1 + tan 45° tan A)
The value of tan 45° is 1.
= (1 - tan A) / (1 + tan A)
Problem 2 :
Prove that cot(A + B) = (cotAcotB − 1)/(cotA + cotB)
Solution :
L.H.S
cot(A + B) = cos (A + B) / sin (A + B)
= (cos A cos B - sin A sin B) / (sin A cos B + cos A sin B)
By dividing every terms by sin A sin B, we get
= (cot A cot B - 1) / (cot B + cot A)
= (cot A cot B - 1) / (cot A + cot B) ---> R.H.S
Hence proved.
Problem 3 :
If tan x = n/(n + 1) and tan y = 1/(2n + 1) , find tan(x + y).
Solution :
tan(x + y) = (tan x + tan y) / (1 - tanx tan y)
= n(2n + 1) + (n + 1) / (n + 1)(2n + 1) - n
= (2n2 + n + n + 1) / (2n2 + n + 2n + 1 - n)
= (2n2 + 2n + 1) / (2n2 + 2n + 1)
tan(x + y) = 1
So, the value of tan(x + y) is 1.
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