Name: Express All Trigonometric Ratios in Term of secA
Uploaded: 2019-07-19
Duration: 5 min 6 s
Description: Learn how to express all the trigonometric ratios in the form of secant A with this comprehensive video on the relationships between trigonometric ratios. Ch...

Question

If $\cos A = \frac{1}{2}$ and $\sin B = \frac{1}{\sqrt{2}}$ , find the value of: $\frac{\tan A - \tan B}{1 + \tan A \tan B}$ . Are angles $A$ and $B$ from the same triangle? Explain.

Accepted Answer

Consider the diagram below:

$\cos A = \frac{1}{2}$

We know that, $\cos A = \frac{base}{hypotenuse}$

$\Rightarrow \frac{A B}{A C} = \frac{1}{2}$

Therefore if length of $A B = x$ , length of $A C = 2 x$

Since, $A B^{2} + B C^{2} = A C^{2}$ [Using Pythagoras Theorem]

$\Rightarrow (x)^{2} + B C^{2} = (2 x)^{2}$

$\Rightarrow B C^{2} = 4 x^{2} - x^{2} = 3 x^{2}$

$\Rightarrow B C = \sqrt{3} x (perpendicular)$

Consider the diagram below:

$\sin B = \frac{1}{\sqrt{2}}$

We know that, $\sin B = \frac{perpendicular}{hypotenuse}$

$\Rightarrow \frac{A C}{B C} = \frac{1}{\sqrt{2}}$

Therefore if length of $A C = x$ , length of $B C = \sqrt{2} x$

Since, $A B^{2} + A C^{2} = B C^{2}$ [Using Pythagoras Theorem]

$\Rightarrow A B^{2} + x^{2} = (\sqrt{2} x)^{2}$

$\Rightarrow A B^{2} = 2 x^{2} - x^{2} = x^{2}$

$∴ A B = x (base)$

Now, $\tan A = \frac{perpendicular}{base} = \frac{\sqrt{3} x}{x} = \sqrt{3}$

$\tan B = \frac{perpendicular}{base} = \frac{x}{x} = 1$

Therefore, $\frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}}$

$= \frac{\sqrt{3} - 1}{1 + \sqrt{3}} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{4 - 2 \sqrt{3}}{2} = 2 - \sqrt{3}$

Thus the value of $\frac{\tan A - \tan B}{1 + \tan A \tan B}$ is $2 - \sqrt{3}$ .

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