Question

if tanA=2, tanB=3, evaluate(cosA.cosB-sinA.sinB)^2

Answer 1

consider two triangles having base angles A & B respectively .

accordingly , in 1st triangle

tanA = P/B=2/1

using Pythagoras theorem , H^2=P^2+B^2

H^2 = 2^2. + 1^2

H^2 = 4+1=5

H=√5.

cosA=base/ hypotenuse=1/√5

sinA=perpendicular/hypotenuse = 2/√5

...

in 2nd triangle

tanB=P/B = 3/1

using PGT

H^2= 3^2 + 1^2

H = √10

cos B = 1/√10

sinB = 3/√10

now( cos A.cosB - sinA .sinB )

= { ( 1/√5)*(1/√10) - (2/√5)*(3/√10) }

={ 1/√50 - 6/√50 }

= -5/√50

= -1/√2