If 3 cot A = 4, check whether (1-tan2 A)/(1+tan2 A) = cos2 A – sin 2 A or not.
Answers
Let us assume a △ABC in which ∠B = 90° and ∠C = θ
Given:
cot θ = BC/AB = 7/8
Let BC = 7k and AB = 8k, where k is a positive real number
According to Pythagoras theorem in △ABC we get.
AC2 = AB2+BC2
AC2 = (8k)2+(7k)2
AC2 = 64k2+49k2
AC2 = 113k2
AC = √113 k
According to the sine and cos function ratios, it is written as
sin θ = AB/AC = Opposite Side/Hypotenuse = 8k/√113 k = 8/√113 and
cos θ = Adjacent Side/Hypotenuse = BC/AC = 7k/√113 k = 7/√113
⇒Solution:-
We have
➠ 3cot A=4
➠cot A=4/3
cot A=base/perpendicular =AB/BC=4/3
Let the sides be in the ratio of k.
We have
AB=4k and BC =3k [as shown in figure]
By pythagorus theorum we have
➠AC²=AB²+BC²
➠AC²=(4k)²+(3k)²
➠AC²=16k²+9k²
➠AC²=25k²
➠AC=5k
Now,
⇛sin A=perpendicular/Hypotenuse=BC/AC=3k/5k=3/5
⇛cos A=base/Hypotenuse =AB/AC=4k/5k=4/5
⇛tan A=perpendicular/base=BC/AB=3k/4k=3/4
We have LHS
⇒ (1-tan²A)/(1+tan²A)
⇒ [1-(3)²/(4)²] [1+(3)²/(4)²]
⇒ [16-9/16][16+9/16]
⇒[7/16]×[16/25]
⇒7/25
And RHS
⇒cos²A-sin²A
⇒ [(4)²/(5)²]-[(3)²/(5)²]
⇒ [16/25]-[9/25]
⇒ 7/25
Therefore
LHS=RHS
