If 3 cot A = 4, check whether 1−tan2 A1+tan2 A=cos2 A−sin2 A or not.

Solution:

Given: $3 \cot A=4$

To check whether $\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$ or not

$3 \cot A=4$

Dividing by 3 on both sides,

We get,

$\cot A=\frac{4}{3}$

By definition,

$\cot A=\frac{1}{\tan A}$

Therefore,

$\cot A=\frac{\frac{1}{\text { Perpendicular side opposite to } \angle \mathrm{A}}}{\text { Base side adjacent to } \angle \mathrm{A}}$

$\cot A=\frac{\text { Base side adjacent to } \angle \mathrm{A}}{\text { Perpendicular side opposite to } \angle \mathrm{A}}$$\ldots \ldots(2)$

Comparing Equation (1) and (2)

We get,

Base side adjacent to $\angle A=4$

Perpendicular side opposite to $\angle A=3$

Hence, $\triangle A B C$ is as shown in figure below

In $\triangle A B C$, Hypotenuse is unknown

Hence, It can be found by using Pythagoras theorem

Therefore by applying Pythagoras theorem in $\triangle A B C$

We get

$A C^{2}=A B^{2}+B C^{2}$

Substituting values of sides from the above figure

$A C^{2}=4^{2}+3^{2}$

$A C^{2}=16+9$

$A C^{2}=25$

$A C=\sqrt{25}$

$A C=5$

Hence, Hypotenuse = 5

To check whether $\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$ or not

We get the values of $\tan A, \cos A, \sin A$

By definition,

$\tan A=\frac{1}{\cot A}$

Substituting the value  from Equation (1)

We get,

$\tan A=\frac{1}{\frac{4}{3}}$

$\tan A=\frac{3}{4}$....(3)

Now by definition,

$\sin A=\frac{\text { Perpendicular side opposite to } \angle A}{\text { Hypotenuse }}$

$\sin A=\frac{B C}{A C}$

Substituting values of sides from the above figure

$\sin A=\frac{3}{5}$.....(4)

Now by definition,

$\cos A=\frac{\text { Base side adjacent to } \angle A}{\text { Hypotenuse }}$

$\cos A=\frac{A B}{A C}$

Substituting values of sides from the above figure

$\cos A=\frac{4}{5} \ldots \ldots$(5)

Now we first take L.H.S of Equation $\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$

L.H.S $=\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$

Substituting value of $\tan A$ from equation (3)

We get,

L.H.S $=\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}$

$\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}$

$\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{1-\left(\frac{3^{2}}{4^{2}}\right)}{1+\left(\frac{3^{2}}{4^{2}}\right)}$

$\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$

Taking L.C.M on both numerator and denominator

We get,

$\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{\frac{16-9}{16}}{\frac{16+9}{16}}$

$\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{7}{25} \ldots \ldots$(6)

Now we take R.H.S of Equation $\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$

R.H.S $=\cos ^{2} A-\sin ^{2} A$

Substituting value of $\sin A$ and $\cos A$ from equation (4) and (5)

We get,

R.H.S $=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}$

$\cos ^{2} A-\sin ^{2} A=\frac{4^{2}}{5^{2}}-\frac{3^{2}}{5^{2}}$

$\cos ^{2} A-\sin ^{2} A=\frac{16}{25}-\frac{9}{25}$

$\cos ^{2} A-\sin ^{2} A=\frac{16-9}{25}$

$\cos ^{2} A-\sin ^{2} A=\frac{7}{25}$....(7)

Comparing Equation (6) and (7)

We get,

$\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$

Answer: Yes $\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A$