CBSE Class 10 Mathematics Chapter 8: Introduction to Trigonometry NCERT Solutions

First, we need to find the length of the hypotenuse AC using the Pythagorean theorem in the right-angled triangle ABC.

$$AC^2\; =\; AB^2\; +\; BC^2$$

Substitute the given values:

$AC^2\; =\; (24\; \backslash text\{\; cm\})^2\; +\; (7\; \backslash text\{\; cm\})^2$

$AC^2\; =\; 576\; \backslash text\{\; cm\}^2\; +\; 49\; \backslash text\{\; cm\}^2$

$AC^2\; =\; 625\; \backslash text\{\; cm\}^2$

Now, take the square root to find AC:

$AC\; =\; \backslash sqrt\{625\}\; \backslash text\{\; cm\}$

$AC\; =\; 25\; \backslash text\{\; cm\}$

Now we can determine the trigonometric ratios:

(i) For angle A:

The side opposite to angle A is BC, and the adjacent side is AB. The hypotenuse is AC.

$\backslash sin\; A\; =\; \backslash frac\{\backslash text\{Side\; opposite\; to\; \}\; \backslash angle\; A\}\{\backslash text\{Hypotenuse\}\}\; =\; \backslash frac\{BC\}\{AC\}\; =\; \backslash frac\{7\; \backslash text\{\; cm\}\}\{25\; \backslash text\{\; cm\}\}\; =\; \backslash frac\{7\}\{25\}$

$\backslash cos\; A\; =\; \backslash frac\{\backslash text\{Side\; adjacent\; to\; \}\; \backslash angle\; A\}\{\backslash text\{Hypotenuse\}\}\; =\; \backslash frac\{AB\}\{AC\}\; =\; \backslash frac\{24\; \backslash text\{\; cm\}\}\{25\; \backslash text\{\; cm\}\}\; =\; \backslash frac\{24\}\{25\}$

(ii) For angle C:

The side opposite to angle C is AB, and the adjacent side is BC. The hypotenuse is AC.

$\backslash sin\; C\; =\; \backslash frac\{\backslash text\{Side\; opposite\; to\; \}\; \backslash angle\; C\}\{\backslash text\{Hypotenuse\}\}\; =\; \backslash frac\{AB\}\{AC\}\; =\; \backslash frac\{24\; \backslash text\{\; cm\}\}\{25\; \backslash text\{\; cm\}\}\; =\; \backslash frac\{24\}\{25\}$

$\backslash cos\; C\; =\; \backslash frac\{\backslash text\{Side\; adjacent\; to\; \}\; \backslash angle\; C\}\{\backslash text\{Hypotenuse\}\}\; =\; \backslash frac\{BC\}\{AC\}\; =\; \backslash frac\{7\; \backslash text\{\; cm\}\}\{25\; \backslash text\{\; cm\}\}\; =\; \backslash frac\{7\}\{25\}$

The given triangle is PQR, right-angled at Q. We are given PQ = 12 cm and PR = 13 cm. We need to find the length of the side QR first.

Using the Pythagorean theorem in triangle PQR:

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

Substitute the given values:

$(13\; \backslash text\{\; cm\})^2\; =\; (12\; \backslash text\{\; cm\})^2\; +\; QR^2$

$169\; \backslash text\{\; cm\}^2\; =\; 144\; \backslash text\{\; cm\}^2\; +\; QR^2$

Rearrange the equation to solve for QR^2:

$QR^2\; =\; 169\; \backslash text\{\; cm\}^2\; -\; 144\; \backslash text\{\; cm\}^2$

$QR^2\; =\; 25\; \backslash text\{\; cm\}^2$

Take the square root to find QR:

$QR\; =\; \backslash sqrt\{25\}\; \backslash text\{\; cm\}$

$QR\; =\; 5\; \backslash text\{\; cm\}$

Now, we need to find tan P and cot R.

For angle P, the opposite side is QR and the adjacent side is PQ.

$\backslash tan\; P\; =\; \backslash frac\{\backslash text\{Opposite\}\}\{\backslash text\{Adjacent\}\}\; =\; \backslash frac\{QR\}\{PQ\}\; =\; \backslash frac\{5\; \backslash text\{\; cm\}\}\{12\; \backslash text\{\; cm\}\}\; =\; \backslash frac\{5\}\{12\}$

For angle R, the opposite side is PQ and the adjacent side is QR.

$\backslash cot\; R\; =\; \backslash frac\{\backslash text\{Adjacent\}\}\{\backslash text\{Opposite\}\}\; =\; \backslash frac\{QR\}\{PQ\}\; =\; \backslash frac\{5\; \backslash text\{\; cm\}\}\{12\; \backslash text\{\; cm\}\}\; =\; \backslash frac\{5\}\{12\}$

Finally, calculate tan P - cot R:

$\backslash tan\; P\; -\; \backslash cot\; R\; =\; \backslash frac\{5\}\{12\}\; -\; \backslash frac\{5\}\{12\}\; =\; 0$

Therefore, the value of tan P - cot R is 0.