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Q8:
If $3 \cot A = 4$, check whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A – \sin^2 A$ or not.

Solution :

Given: $3 \cot A = 4$

To Check: Whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$

A B C 4k 3k 5k

Step 1: Determine the sides of the right-angled triangle.

Given $3 \cot A = 4$, we can write $\cot A = \frac{4}{3}$.

In a right-angled triangle, $\cot A = \frac{\text{Adjacent side}}{\text{Opposite side}} = \frac{AB}{BC}$.

Let $AB = 4k$ and $BC = 3k$, where $k$ is a positive constant.

Using the Pythagoras Theorem ($AC^2 = AB^2 + BC^2$):

$AC^2 = (4k)^2 + (3k)^2$

$AC^2 = 16k^2 + 9k^2 = 25k^2$

$AC = \sqrt{25k^2} = 5k$

Step 2: Calculate the trigonometric ratios.

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

$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{3k}{5k} = \frac{3}{5}$

$\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5}$

Step 3: Evaluate the Left Hand Side (LHS).

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

Substitute $\tan A = \frac{3}{4}$:

LHS = $\frac{1 - (\frac{3}{4})^2}{1 + (\frac{3}{4})^2} = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}}$

LHS = $\frac{\frac{16-9}{16}}{\frac{16+9}{16}} = \frac{\frac{7}{16}}{\frac{25}{16}} = \frac{7}{25}$

Step 4: Evaluate the Right Hand Side (RHS).

RHS = $\cos^2 A - \sin^2 A$

Substitute $\cos A = \frac{4}{5}$ and $\sin A = \frac{3}{5}$:

RHS = $(\frac{4}{5})^2 - (\frac{3}{5})^2$

RHS = $\frac{16}{25} - \frac{9}{25} = \frac{16 - 9}{25} = \frac{7}{25}$

Step 5: Conclusion.

Since LHS = $\frac{7}{25}$ and RHS = $\frac{7}{25}$, the equation holds true.

Final Answer: Yes, $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ is true.


More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.1


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