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Introduction to Trigonometry

Madhya Pradesh Board · Class 10 · Mathematics

NCERT Solutions for Introduction to Trigonometry — Madhya Pradesh Board Class 10 Mathematics.

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19 Questions Solved · 3 Sections

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EXERCISE 8.1

1In ΔABC\Delta ABC, right-angled at B, AB=24AB = 24 cm, BC=7BC = 7 cm. Determine :Show solution
In right triangle ABCABC right-angled at BB:

- Hypotenuse AC=AB2+BC2=242+72=576+49=625=25AC = \sqrt{AB^2+BC^2} = \sqrt{24^2+7^2} = \sqrt{576+49} = \sqrt{625} = 25 cm.

For angle AA:
- sin⁡A=BCAC=725\sin A = \dfrac{BC}{AC} = \dfrac{7}{25}
- cos⁡A=ABAC=2425\cos A = \dfrac{AB}{AC} = \dfrac{24}{25}

For angle CC:
- sin⁡C=ABAC=2425\sin C = \dfrac{AB}{AC} = \dfrac{24}{25}
- cos⁡C=BCAC=725\cos C = \dfrac{BC}{AC} = \dfrac{7}{25}

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2In Fig. 8.13, find tan⁡P−cot⁡R\tan P - \cot R.Show solution
From Fig. 8.13, the required expression simplifies to the difference of two equal trigonometric ratios for the same acute angles. Using the values from the figure, tan⁡P=cot⁡R\tan P = \cot R, so

tan⁡P−cot⁡R=0.\tan P - \cot R = 0.

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

Using sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1,

cos⁡2A=1−sin⁡2A=1−(34)2=1−916=716\cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{3}{4}\right)^2 = 1 - \frac{9}{16} = \frac{7}{16}

Since AA is acute,

cos⁡A=74\cos A = \frac{\sqrt{7}}{4}

Now,

tan⁡A=sin⁡Acos⁡A=3/47/4=37=377\tan A = \frac{\sin A}{\cos A} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}} = \frac{3\sqrt{7}}{7}

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4Given 15cot⁡A=815 \cot A = 8, find sin⁡A\sin A and sec⁡A\sec A.Show solution
Given 15cot⁡A=815\cot A=8, so

cot⁡A=815\cot A=\frac{8}{15}

Take adjacent : opposite =8:15=8:15.

Then hypotenuse

=82+152=64+225=289=17=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17

So,

sin⁡A=1517\sin A=\frac{15}{17}

and

sec⁡A=178\sec A=\frac{17}{8}

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5Given sec⁡θ=1312\sec \theta = \frac{13}{12}, calculate all other trigonometric ratios.Show solution
Given sec⁡θ=1312\sec\theta=\dfrac{13}{12}, so

cos⁡θ=1213\cos\theta=\frac{12}{13}

Using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1,

sin⁡2θ=1−(1213)2=1−144169=25169\sin^2\theta=1-\left(\frac{12}{13}\right)^2=1-\frac{144}{169}=\frac{25}{169}

Since θ\theta is acute,

sin⁡θ=513\sin\theta=\frac{5}{13}

Now,

tan⁡θ=sin⁡θcos⁡θ=5/1312/13=512\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{5/13}{12/13}=\frac{5}{12}

cot⁡θ=125,csc⁡θ=135\cot\theta=\frac{12}{5},\qquad \csc\theta=\frac{13}{5}

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6If ∠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.Show solution
For acute angles, cosine is one-to-one.

Given cos⁡A=cos⁡B\cos A = \cos B and both AA and BB are acute, the angles must be equal.

So, ∠A=∠B\angle A = \angle B.

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7If cot⁡θ=78\cot \theta = \frac{7}{8}, evaluate :Show solution
Given cot⁡θ=78\cot\theta=\dfrac{7}{8}.

Take adjacent : opposite =7:8=7:8.
Then hypotenuse

=72+82=49+64=113=\sqrt{7^2+8^2}=\sqrt{49+64}=\sqrt{113}

So,

sin⁡θ=8113,cos⁡θ=7113\sin\theta=\frac{8}{\sqrt{113}},\qquad \cos\theta=\frac{7}{\sqrt{113}}

(i)

(1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)=1−sin⁡2θ1−cos⁡2θ=cos⁡2θsin⁡2θ=cot⁡2θ=(78)2=4964\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)} =\frac{1-\sin^2\theta}{1-\cos^2\theta} =\frac{\cos^2\theta}{\sin^2\theta} =\cot^2\theta =\left(\frac{7}{8}\right)^2=\frac{49}{64}

(ii)

cot⁡2θ=(78)2=4964\cot^2\theta=\left(\frac{7}{8}\right)^2=\frac{49}{64}

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8If 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.Show solution
Given 3cot⁡A=43\cot A=4, so

cot⁡A=43\cot A=\frac{4}{3}

Hence,

tan⁡A=34\tan A=\frac{3}{4}

Now,

1−tan⁡2A1+tan⁡2A=1−(34)21+(34)2=1−9161+916=7162516=725\frac{1-\tan^2 A}{1+\tan^2 A}=\frac{1-\left(\frac{3}{4}\right)^2}{1+\left(\frac{3}{4}\right)^2} =\frac{1-\frac{9}{16}}{1+\frac{9}{16}} =\frac{\frac{7}{16}}{\frac{25}{16}}=\frac{7}{25}

Also, using a 33-44-55 triangle, if opposite =3=3, adjacent =4=4, then hypotenuse =5=5.
So

sin⁡A=35,cos⁡A=45\sin A=\frac{3}{5},\quad \cos A=\frac{4}{5}

Therefore,

cos⁡2A−sin⁡2A=(45)2−(35)2=16−925=725\cos^2 A-\sin^2 A=\left(\frac{4}{5}\right)^2-\left(\frac{3}{5}\right)^2=\frac{16-9}{25}=\frac{7}{25}

Thus,

1−tan⁡2A1+tan⁡2A=cos⁡2A−sin⁡2A\frac{1-\tan^2 A}{1+\tan^2 A}=\cos^2 A-\sin^2 A

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9In triangle ABC, right-angled at B, if tan⁡A=13\tan A = \frac{1}{\sqrt{3}}, find the value of:Show solution
If tan⁡A=13\tan A=\dfrac{1}{\sqrt{3}}, then from the table of special angles,

A=30∘A=30^\circ

Since C=60∘C=60^\circ in a right triangle,

(i)

sin⁡Acos⁡C+cos⁡Asin⁡C=sin⁡(A+C)=sin⁡90∘=1\sin A\cos C+\cos A\sin C=\sin(A+C)=\sin 90^\circ=1

(ii)

cos⁡Acos⁡C−sin⁡Asin⁡C=cos⁡(A+C)=cos⁡90∘=0\cos A\cos C-\sin A\sin C=\cos(A+C)=\cos 90^\circ=0

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10In ΔPQR\Delta PQR, right-angled at Q, PR+QR=25PR + QR = 25 cm and PQ=5PQ = 5 cm. Determine the values of sin⁡P\sin P, cos⁡P\cos P and tan⁡P\tan P.Show solution
In right triangle PQRPQR, right-angled at QQ.
Let QR=xQR=x. Given PR+QR=25PR+QR=25 and PQ=5PQ=5.

Using Pythagoras:

PR2=PQ2+QR2=52+x2PR^2=PQ^2+QR^2=5^2+x^2

and

PR=25−xPR=25-x

So,

(25−x)2=25+x2(25-x)^2=25+x^2

625−50x+x2=25+x2625-50x+x^2=25+x^2

600=50x600=50x

x=12x=12

Hence QR=12QR=12 and PR=13PR=13.

For angle PP:

sin⁡P=QRPR=1213,cos⁡P=PQPR=513,tan⁡P=QRPQ=125\sin P=\frac{QR}{PR}=\frac{12}{13},\quad \cos P=\frac{PQ}{PR}=\frac{5}{13},\quad \tan P=\frac{QR}{PQ}=\frac{12}{5}

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11State whether the following are true or false. Justify your answer.

EXERCISE 8.2

1Evaluate the following :
2Choose the correct option and justify your choice :
3If tan⁡(A+B)=3\tan (A + B) = \sqrt{3} and tan⁡(A−B)=13\tan (A - B) = \frac{1}{\sqrt{3}} ; 0∘<A+B≤90∘0^\circ < A + B \leq 90^\circ ; A>BA > B , find A and B.
4State whether the following are true or false. Justify your answer.

EXERCISE 8.3

1Express the trigonometric ratios sin⁡A\sin A, sec⁡A\sec A and tan⁡A\tan A in terms of cot⁡A\cot A.
2Write all the other trigonometric ratios of ∠A\angle A in terms of sec⁡A\sec A.
3Choose the correct option. Justify your choice.
4Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

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