Solution:

In ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(i) sin A, cos A:

In right-angled triangle ABC, right-angled at B:

AC = $\sqrt{AB^2 + BC^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$ cm

Now, for angle A:

sin A = $\frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$

cos A = $\frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$

(ii) sin C, cos C:

For angle C:

sin C = $\frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$

cos C = $\frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$

Solution:
QR = $\sqrt{13^2 - 12^2}$ = $\sqrt{169 - 144}$ = 5 cm

tan P = $\dfrac{5}{12}$

cot R = $\dfrac{5}{12}$

tan P - cot R = 0

Solution:

Given: sin A = $\dfrac{3}{4}$

To calculate cos A and tan A:

Step 1: Use the identity $\sin^2 A + \cos^2 A = 1$ to find cos A.

We know that $\sin A = \dfrac{3}{4}$, so:

$\sin^2 A = \left( \dfrac{3}{4} \right)^2 = \dfrac{9}{16}$

Substitute into the identity:

$\dfrac{9}{16} + \cos^2 A = 1$

Now solve for $\cos^2 A$:

$\cos^2 A = 1 - \dfrac{9}{16} = \dfrac{16}{16} - \dfrac{9}{16} = \dfrac{7}{16}$

Taking the square root of both sides:

$\cos A = \pm \dfrac{\sqrt{7}}{4}$

Step 2: To find tan A, use the identity $\tan A = \dfrac{\sin A}{\cos A}$.

We know $\sin A = \dfrac{3}{4}$ and $\cos A = \pm \dfrac{\sqrt{7}}{4}$, so:

$\tan A = \dfrac{\dfrac{3}{4}}{\pm \dfrac{\sqrt{7}}{4}} = \pm \dfrac{3}{\sqrt{7}}$

Rationalizing the denominator:

$\tan A = \pm \dfrac{3}{\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}} = \pm \dfrac{3\sqrt{7}}{7}$

Solution:

Given: 15 cot A = 8

Step 1: Express cot A in terms of sin A and cos A

We know that cot A = $\frac{\cos A}{\sin A}$. Therefore, we can write:

15 $\cdot$ $\frac{\cos A}{\sin A}$ = 8

Step 2: Solve for $\cos A$ in terms of $\sin A$

$\frac{\cos A}{\sin A}$ = $\frac{8}{15}$

So, $\cos A$ = $\frac{8}{15}$ $\cdot$ $\sin A$

Step 3: Use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ to find $\sin A$

Substitute $\cos A$ = $\frac{8}{15}$ $\cdot$ $\sin A$ into the identity:

$(\sin^2 A) + \left(\frac{8}{15} \cdot \sin A\right)^2 = 1$

$(\sin^2 A) + \frac{64}{225} \cdot \sin^2 A = 1$

Step 4: Factor the equation

Factor out $\sin^2 A$:

$\sin^2 A \left(1 + \frac{64}{225}\right) = 1$

Step 5: Simplify

$1 + \frac{64}{225}$ = $\frac{225 + 64}{225}$ = $\frac{289}{225}$

So the equation becomes:

$\sin^2 A \cdot \frac{289}{225} = 1$

Step 6: Solve for $\sin^2 A$

$\sin^2 A = \frac{225}{289}$

Step 7: Find $\sin A$

$\sin A = \frac{15}{17}$

Step 8: Find $\cos A$

We know that $\cos^2 A = 1 - \sin^2 A$, so:

$\cos^2 A = 1 - \frac{225}{289}$ = $\frac{64}{289}$

$\cos A = \frac{8}{17}$

Step 9: Find sec A

We know that sec A = $\frac{1}{\cos A}$, so:

$\sec A = \frac{1}{\frac{8}{17}} = \frac{17}{8}$

Solution:

Given sec $\theta$ = $\dfrac{13}{12}$, calculate all other trigonometric ratios

We are given that sec $\theta$ = $\dfrac{13}{12}$.

We know that:

• sec $\theta$ = $\dfrac{1}{\cos \theta}$
• cos $\theta$ = $\dfrac{12}{13}$

Now, let's calculate the other trigonometric ratios.

1. To find sin $\theta$, we use the Pythagorean identity:

• sin$^2 \theta$ + cos$^2 \theta$ = 1
• sin$^2 \theta$ = 1 - cos$^2 \theta$ = 1 - $\left(\dfrac{12}{13}\right)^2$ = 1 - $\dfrac{144}{169}$ = $\dfrac{25}{169}$
• sin $\theta$ = $\dfrac{5}{13}$ (taking the positive root assuming $\theta$ is in the first quadrant)

2. To find tan $\theta$, we use the formula:

• tan $\theta$ = $\dfrac{\sin \theta}{\cos \theta}$
• tan $\theta$ = $\dfrac{\dfrac{5}{13}}{\dfrac{12}{13}}$ = $\dfrac{5}{12}$

3. To find cot $\theta$, we use the formula:

• cot $\theta$ = $\dfrac{1}{\tan \theta}$
• cot $\theta$ = $\dfrac{12}{5}$

The trigonometric ratios are:

• sin $\theta$ = $\dfrac{5}{13}$
• cos $\theta$ = $\dfrac{12}{13}$
• tan $\theta$ = $\dfrac{5}{12}$
• cot $\theta$ = $\dfrac{12}{5}$
• sec $\theta$ = $\dfrac{13}{12}$

Solution:

Given:

Let $\angle A$ and $\angle B$ be acute angles such that $\cos A = \cos B$.

To Prove:

We need to prove that $\angle A = \angle B$.

Proof:

Since $\angle A$ and $\angle B$ are acute angles, they lie in the range $0^\circ < A, B < 90^\circ$.

We are given that $\cos A = \cos B$.

We know that the cosine function is one-to-one in the range of acute angles (i.e., $0^\circ < \theta < 90^\circ$). This implies that if $\cos A = \cos B$, then $A = B$.

Therefore, $\angle A = \angle B$.

Solution:

In triangle ABC, right-angled at B, if tan A = $\dfrac{1}{\sqrt{3}}$, find the value of:

Solution:

Given:

We are given that $3 \cot A = 4$.

Step 1: Express $\cot A$ in terms of $A$

From the given equation, $3 \cot A = 4$, we can solve for $\cot A$:

$\cot A = \dfrac{4}{3}$

Step 2: Use the identity for $\cot A$

We know that $\cot A = \dfrac{\cos A}{\sin A}$, so:

$\dfrac{\cos A}{\sin A} = \dfrac{4}{3}$

Thus, $\cos A = \dfrac{4}{3} \sin A$.

Step 3: Use the Pythagorean identity

We know that $\cos^2 A + \sin^2 A = 1$. Substituting $\cos A = \dfrac{4}{3} \sin A$ into this identity:

$(\dfrac{4}{3} \sin A)^2 + \sin^2 A = 1$

$\dfrac{16}{9} \sin^2 A + \sin^2 A = 1$

Factoring out $\sin^2 A$:

$(\dfrac{16}{9} + 1) \sin^2 A = 1$

$(\dfrac{16}{9} + \dfrac{9}{9}) \sin^2 A = 1$

$(\dfrac{25}{9}) \sin^2 A = 1$

So, $\sin^2 A = \dfrac{9}{25}$.

Step 4: Find $\cos^2 A$

Since $\cos^2 A + \sin^2 A = 1$, we can now find $\cos^2 A$:

$\cos^2 A = 1 - \sin^2 A = 1 - \dfrac{9}{25} = \dfrac{25}{25} - \dfrac{9}{25} = \dfrac{16}{25}$.

Step 5: Simplify the given expression

We are asked to check whether $\dfrac{1 - \tan^2 A}{1 + \tan^2 A} - \cos^2 A - \sin^2 A$ is true. We will simplify each part.

Step 6: Use the identity for $\tan A$

We know that $\tan A = \dfrac{1}{\cot A}$, so $\tan A = \dfrac{3}{4}$.

Now, $\tan^2 A = \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}$.

Step 7: Simplify the first part of the expression

We need to simplify $\dfrac{1 - \tan^2 A}{1 + \tan^2 A}$:

$\dfrac{1 - \dfrac{9}{16}}{1 + \dfrac{9}{16}} = \dfrac{\dfrac{16}{16} - \dfrac{9}{16}}{1 + \dfrac{9}{16}} = \dfrac{\dfrac{7}{16}}{\dfrac{25}{16}} = \dfrac{7}{25}$.

Step 8: Final simplification

Now, we substitute into the original expression:

$\dfrac{7}{25} - \cos^2 A - \sin^2 A = \dfrac{7}{25} - \dfrac{16}{25} - \dfrac{9}{25}$

$= \dfrac{7 - 16 - 9}{25} = \dfrac{-18}{25}$.

Conclusion:

The value of the expression is $\dfrac{-18}{25}$.

Solution:

Given:

In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm.

To find:

We need to determine the values of $ \sin P $, $ \cos P $, and $ \tan P $.

Step 1: Use the Pythagorean theorem to find PR.

Since the triangle is right-angled at Q, we can apply the Pythagorean theorem:

$ PR^2 = PQ^2 + QR^2 $

Substituting the known values:

$ PR^2 = 5^2 + QR^2 $

$ PR^2 = 25 + QR^2 $

Step 2: Use the given condition PR + QR = 25 cm.

From the given condition, we have:

$ PR + QR = 25 $

Let PR = x and QR = y, then:

$ x + y = 25 $

Also, from the Pythagorean theorem, we know:

$ x^2 = 25 + y^2 $

Step 3: Solve the system of equations.

From $ x + y = 25 $, we can express x as:

$ x = 25 - y $

Substitute this value of x into $ x^2 = 25 + y^2 $:

$ (25 - y)^2 = 25 + y^2 $

Expanding the left side:

$ 625 - 50y + y^2 = 25 + y^2 $

Cancel $ y^2 $ from both sides:

$ 625 - 50y = 25 $

$ 600 = 50y $

$ y = 12 $

Step 4: Find x (PR).

Now substitute $ y = 12 $ into $ x = 25 - y $:

$ x = 25 - 12 = 13 $

Step 5: Calculate sin P, cos P, and tan P.

Now, we can calculate the trigonometric ratios for angle P:

1. $ \sin P = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{QR}{PR} = \frac{12}{13} $

2. $ \cos P = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PQ}{PR} = \frac{5}{13} $

3. $ \tan P = \frac{\text{opposite}}{\text{adjacent}} = \frac{QR}{PQ} = \frac{12}{5} $

Solution:

(i) The value of tan A is always less than 1.

False. The value of $tan A$ can be greater than 1 for certain angles. For example, at $A = 45^\circ$, $tan A = 1$, and for $A = 60^\circ$, $tan A > 1$.

(ii) sec A = 12 for some value of angle A.

True. $sec A = 12$ is possible for some angle $A$. Since $sec A = \dfrac{1}{cos A}$, for $sec A = 12$, $cos A = \dfrac{1}{12}$, which is a valid value for cosine of an angle.

(iii) cos A is the abbreviation used for the cosecant of angle A.

False. $cos A$ refers to the cosine of angle $A$, not the cosecant. The cosecant is abbreviated as $cosec A$.

(iv) cot A is the product of cot and A.

False. $cot A$ is the reciprocal of $tan A$, i.e., $cot A = \dfrac{1}{tan A}$. It is not the product of $cot$ and $A$.

(v) sin $\theta$ = $\dfrac{4}{3}$ for some angle $\theta$.

False. The value of $sin \theta$ lies between -1 and 1 for all real angles. $\dfrac{4}{3}$ is not a valid value for $sin \theta$.