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Q9(ii):
In triangle ABC, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of: (ii) $\cos A \cos C – \sin A \sin C$

Solution :

Given: In $\triangle ABC$, $\angle B = 90^\circ$ and $\tan A = \frac{1}{\sqrt{3}}$.

To find: The value of $\cos A \cos C - \sin A \sin C$.

B C A $\sqrt{3}k$ $k$ $2k$

Step 1: Determine the sides of the triangle.

By definition of the tangent ratio in a right-angled triangle, $\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AB}$.

Given $\tan A = \frac{1}{\sqrt{3}}$, let $BC = k$ and $AB = \sqrt{3}k$, where $k$ is a positive constant.

Step 2: Calculate the hypotenuse $AC$ using the Pythagoras Theorem.

The Pythagoras Theorem states: $AC^2 = AB^2 + BC^2$.

$AC^2 = (\sqrt{3}k)^2 + (k)^2$

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

$AC = \sqrt{4k^2} = 2k$

Step 3: Determine the trigonometric ratios for angles A and C.

For angle $A$: Opposite side = $BC = k$, Adjacent side = $AB = \sqrt{3}k$, Hypotenuse = $AC = 2k$.

$\sin A = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}$

$\cos A = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}$

For angle $C$: Opposite side = $AB = \sqrt{3}k$, Adjacent side = $BC = k$, Hypotenuse = $AC = 2k$.

$\sin C = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}$

$\cos C = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}$

Step 4: Evaluate the expression $\cos A \cos C - \sin A \sin C$.

Substitute the values calculated in Step 3:

Expression $= (\frac{\sqrt{3}}{2} \times \frac{1}{2}) - (\frac{1}{2} \times \frac{\sqrt{3}}{2})$

Expression $= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}$

Expression $= 0$

Final Answer: 0


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