Let's go through the questions from the image one by one. I'll solve them in detail.

Question 1:

In $\triangle ABC$, right angle at $B$, $AB = 12 \text{ cm}$, and $BC = 5 \text{ cm}$. Find the values of $\sin A$ and $\cos A$ and $\tan A$ and $\sin C$, $\cos C$, and $\tan C$.

Given:

• $AB = 12 \text{ cm}$
• $BC = 5 \text{ cm}$
• $\angle B = 90^\circ$

Using the Pythagorean theorem:

$AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}$

Now, calculate the trigonometric ratios:

For $\angle A$:

$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{5}{13}$ $\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{12}{13}$ $\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB} = \frac{5}{12}$

For $\angle C$:

$\sin C = \frac{AB}{AC} = \frac{12}{13}$ $\cos C = \frac{BC}{AC} = \frac{5}{13}$ $\tan C = \frac{AB}{BC} = \frac{12}{5}$

Question 2:

Given $\sin \theta = \frac{3}{5}$, find all other trigonometric ratios.

Since $\sin \theta = \frac{3}{5}$, let's assume a right triangle where:

• Opposite side = 3
• Hypotenuse = 5

To find the adjacent side, use the Pythagorean theorem:

$\text{Adjacent side} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$

Now, calculate the other trigonometric ratios:

$\cos \theta = \frac{4}{5}, \quad \tan \theta = \frac{3}{4}$ $\csc \theta = \frac{1}{\sin \theta} = \frac{5}{3}, \quad \sec \theta = \frac{1}{\cos \theta} = \frac{5}{4}, \quad \cot \theta = \frac{1}{\tan \theta} = \frac{4}{3}$

Question 3:

If $\sin A = \frac{4}{5}$, evaluate $\frac{1 - \sin^2 A}{\cos^2 A}$.

Given $\sin A = \frac{4}{5}$, we have:

$\sin^2 A = \left(\frac{4}{5}\right)^2 = \frac{16}{25}$

So,

$1 - \sin^2 A = 1 - \frac{16}{25} = \frac{9}{25}$

( \cos^2 A =