if a+b=45 prove that (1+tana)(1+tanb)=2

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)