If 3 cot A = 4, check whether \(\dfrac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A\) or not.
Yes
Given: \(3\cot A = 4\).
To check: whether
\(\dfrac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A\).
Step 1: Find \(\cot A\) and \(\tan A\)
From \(3\cot A = 4\), divide both sides by 3:
\(\cot A = \dfrac{4}{3}\).
Since \(\tan A = \dfrac{1}{\cot A}\), we get:
\(\tan A = \dfrac{3}{4}\).
Step 2: Check the LHS
LHS = \(\dfrac{1-\tan^2 A}{1+\tan^2 A}\).
First square \(\tan A\):
\(\tan^2 A = \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}\).
Now substitute:
\(\text{LHS} = \dfrac{1-\frac{9}{16}}{1+\frac{9}{16}}\)
Convert 1 into \(\dfrac{16}{16}\):
\(\text{LHS} = \dfrac{\frac{16}{16}-\frac{9}{16}}{\frac{16}{16}+\frac{9}{16}} = \dfrac{\frac{7}{16}}{\frac{25}{16}}\)
Divide the fractions:
\(\text{LHS} = \frac{7}{16} \times \frac{16}{25} = \dfrac{7}{25}\).
Step 3: Find \(\sin A\) and \(\cos A\) using a right triangle idea
We have \(\tan A = \dfrac{3}{4}\). Think of a right triangle where:
Opposite side = 3 units, Adjacent side = 4 units.
Then hypotenuse = \(\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\) units.
Student Note: This is the standard \(3\text{-}4\text{-}5\) right triangle, which makes \(\sin\) and \(\cos\) very easy to write.
So,
\(\sin A = \dfrac{3}{5}\), and \(\cos A = \dfrac{4}{5}\).
Step 4: Check the RHS
RHS = \(\cos^2 A - \sin^2 A\).
\(\cos^2 A = \left(\dfrac{4}{5}\right)^2 = \dfrac{16}{25}\)
\(\sin^2 A = \left(\dfrac{3}{5}\right)^2 = \dfrac{9}{25}\)
So,
\(\text{RHS} = \dfrac{16}{25} - \dfrac{9}{25} = \dfrac{7}{25}\).
Step 5: Compare LHS and RHS
LHS = \(\dfrac{7}{25}\) and RHS = \(\dfrac{7}{25}\).
Since LHS = RHS, the given relation is true.
Conclusion: Yes, \(\dfrac{1-\tan^2 A}{1+\tan^2 A} = \cos^2 A - \sin^2 A\) is verified for \(3\cot A = 4\).