Introduction to Trigonometry | Exercise 8.1

Question 2

In Fig. 8.13, find tan⁡P−cot⁡R\tan P - \cot R.

Question diagram 1
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Solution

Right-Angled Triangle: A triangle with one angle equal to 90°. The side opposite the right angle is the longest side, called the hypotenuse. The other two sides are called the legs.

Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse equals the sum of squares of the other two sides:

hypotenuse2=side12+side22\text{hypotenuse}^2 = \text{side}_1^2 + \text{side}_2^2

Here, PR is the hypotenuse. We first find the missing side QR using Pythagoras, then calculate tan P and cot R.

We need to find the length of the unknown side first.

Step 1 — Find side QR

Let's use the Pythagoras theorem in △PQR\triangle PQR. The hypotenuse is PR.

PR2=PQ2+QR2PR^2 = PQ^2 + QR^2

132=122+QR213^2 = 12^2 + QR^2

169=144+QR2169 = 144 + QR^2

QR2=169−144QR^2 = 169 - 144

QR2=25QR^2 = 25

QR=25QR = \sqrt{25}

QR=5 cm\boxed{QR = 5 \text{ cm}}

Diagram 1

Step 2 — Calculate tan P

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

tan⁡P=OppositeAdjacent\tan P = \frac{\text{Opposite}}{\text{Adjacent}}

tan⁡P=QRPQ\tan P = \frac{QR}{PQ}

tan⁡P=512\tan P = \frac{5}{12}

tan⁡P=512\boxed{\tan P = \frac{5}{12}}

Step 3 — Calculate cot R

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

cot⁡R=AdjacentOpposite\cot R = \frac{\text{Adjacent}}{\text{Opposite}}

cot⁡R=QRPQ\cot R = \frac{QR}{PQ}

cot⁡R=512\cot R = \frac{5}{12}

cot⁡R=512\boxed{\cot R = \frac{5}{12}}

Step 4 — Find the difference

Now, we will find the value of tan⁡P−cot⁡R\tan P - \cot R.

tan⁡P−cot⁡R=512−512\tan P - \cot R = \frac{5}{12} - \frac{5}{12}

=0= 0

tan⁡P−cot⁡R=0\boxed{\tan P - \cot R = 0}

Answer

(i) tan⁡P−cot⁡R=0\tan P - \cot R = 0

More questions in Exercise 8.1

Q1

In ΔABC\Delta \text{ABC}, right-angled at B, AB=24 cm\text{AB} = 24\text{ cm}, BC=7 cm\text{BC} = 7\text{ cm}. Determine :

(i) sin⁡A\sin A, cos⁡A\cos A (ii) sin⁡C\sin C, cos⁡C\cos C

Q2

In Fig. 8.13, find tan⁡P−cot⁡R\tan P - \cot R.

Q3

If sin⁡A=34\sin A = \frac{3}{4}, calculate cos⁡A\cos A and tan⁡A\tan A.

Q4

Given 15cot⁡A=815 \cot A = 8, find sin⁡A\sin A and sec⁡A\sec A.

Q5

Given sec⁡θ=1312\sec \theta = \frac{13}{12}, calculate all other trigonometric ratios.

Q6

If ∠A\angle A and ∠B\angle B are acute angles such that cos⁡A=cos⁡B\cos A = \cos B, then show that ∠A=∠B\angle A = \angle B.

Q7

If cot⁡θ=78\cot \theta = \frac{7}{8}, evaluate :

(i) (1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)} (ii) cot⁡2θ\cot^2 \theta

Q8

If 3cot⁡A=43 \cot A = 4, check whether 1−tan⁡2A1+tan⁡2A=cos⁡2A−sin⁡2A\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A or not.

Q9

In triangle ABC, right-angled at B, if tan⁡A=13\tan A = \frac{1}{\sqrt{3}}, find the value of:

(i) sin⁡Acos⁡C+cos⁡Asin⁡C\sin A \cos C + \cos A \sin C

(ii) cos⁡Acos⁡C−sin⁡Asin⁡C\cos A \cos C - \sin A \sin C

Q10

In ΔPQR\Delta \text{PQR}, right-angled at Q, PR+QR=25 cm\text{PR} + \text{QR} = 25\text{ cm} and PQ=5 cm\text{PQ} = 5\text{ cm}. Determine the values of sin⁡P\sin P, cos⁡P\cos P and tan⁡P\tan P.

Q11

State whether the following are true or false. Justify your answer.

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

(ii) sec⁡A=125\sec A = \frac{12}{5} for some value of angle A.

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

(iv) cot⁡A\cot A is the product of cot⁡\cot and AA.

(v) sin⁡θ=43\sin \theta = \frac{4}{3} for some angle θ\theta.

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