if a+b=π/4 prove that (1+tana)(1+tanb)=2
Solution:-
Given that a+b=45
we take tan both side
⇒tan(a+b)=tan45°
⇒(tana + tanb)/(1-tana .tanb )=1
cross multiplication,
⇒(tana + tanb)=1-tana .tanb
adding both side 1 in LHS and RHS
⇒(tana + tanb)+1=1-tana .tanb+1
⇒(tana + tanb+1+tana .tanb=1+1
⇒tana + 1+tanb+tana .tanb=2
⇒(1+tana)+tanb+tana .tanb=2
⇒(1+tana)+tanb(1+tana )=2
⇒(1+tana)(1+tanb)=2
proved and
if a+b=45° prove that (cota-1)(cotb-1)=2
Solution:-
Given that a+b=45
we take cot both side
⇒cot(a+b)=cot45°
⇒(cotb.cota-1)/(cotb + cota)=1
cross multiplication,
⇒(cotb.cota-1)=(cotb + cota)
adding both side -1 in LHS and RHS
⇒cotb.cota-1-1=cotb+ (cota-1)
⇒cotb.cota -cotb- (cota-1)=2
⇒cotb(cota -1)-1 (cota-1)=2
take common in LHS side (cota -1)
⇒(cotb-1) (cota-1)=2
proved.
If A + B = 45°, show that (1 + tan A) (1 + tan B) = 2 and hence deduce the value of tan 22(1/2)°
Given that A+B=45
we take tan both side
⇒tan(A+B)=tan45°
⇒(tanA + tanB)/(1-tanA .tanB )=1
cross multiplication,
⇒(tanA + tanB)=1-tanA .tanB
adding both side 1 in LHS and RHS
⇒(tanA + tanB)+1=1-tanA.tanB+1
⇒(tanA + tanB+1+tanA.tanB=1+1
⇒tanA+ 1+tanB+tanA.tanB=2
⇒(1+tanA)+tanB+tanA .tanB=2
⇒(1+tanA)+tanB(1+tanA )=2
⇒(1+tanA)(1+tanB)=2
proved
and tan 22(1/2)° =
let θ= 22(1/2)°
or θ= (45/2)°
Multiplying both side by 2.
2θ=45°
take tan both side
⇒tan2θ= tan45°
⇒2tanθ/1-tan^2θ=1
cross multiplication,
⇒2tanθ=1-tan^2θ
⇒tan^2θ+2tanθ-1=0 ..............(1)
tanθ=[-2 ±√{(2)^2 - 4.(1).(-1)}]/2
tanθ=-2 ±√(4+4)/2
tanθ=[-2 ±√8]/2
tanθ=[-2 ±2√2]/2
tanθ= -1 ±√2
put value of theta
tan 22(1/2)° =-1 ±√2
The value of tan22(1/2)° only postive beacuse 22(1/2)° is in first Quadrant so in first quadrant all trigonometric ratio is postive also tan22(1/2)°
tan 22(1/2)° =√2 - 1 answer
if a-b=π/4 then find (1+tana)(1-tanb)
(1+tana)(1+tanb) formula
if a-b=45 prove that (1+tana)(1+tanb)
