NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Students who wish to access the NCERT solutions for class 10 chapter 8 can click on the link below to download the entire solution in PDF.

Download PDF

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Exercises)

Q1. In $\Delta \: ABC$ , right-angled at $B, AB = 24 \: cm$ , $BC = 7 \: cm$.
Determine : $(i)\; \sin A, \cos A$ $(ii)\; \sin C, \cos C$

Answer:

We have,
In $\Delta \: ABC$ , $\angle$B = 90, and the length of the base (AB) = 24 cm and length of perpendicular (BC) = 7 cm
So, by using Pythagoras' theorem,
$AC^2 = AB^2 + BC^2$
⇒ $AC = \sqrt{AB^2+BC^2}$
⇒ $AC = \sqrt{576+49}$
⇒ $AC = \sqrt{625}$
⇒ $AC = 25$ cm
Now,
(i) $\sin A = \frac{P}{H} = \frac{BC}{AC} = \frac{7}{25}$
$\cos A = \frac{B}{H} = \frac{BA}{AC} = \frac{24}{25}$

(ii) For angle C, AB is perpendicular to the base (BC).
Here, B indicates Base and P means perpendicular wrt angle $\angle$C
So, $\sin C = \frac{P}{H} = \frac{BA}{AC} = \frac{24}{25}$
and $\cos C = \frac{B}{H} = \frac{BC}{AC} = \frac{7}{25}$

Q2. In Fig. 8.13, find $\tan P - \cot R$ .

Answer:
We have $\Delta$PQR
According to Pythagoras, in a right-angled triangle, the lengths of PQ and PR are 12 cm and 13 cm, respectively.
So, by using Pythagoras' theorem,
$QR = \sqrt{13^2-12^2}$
$QR = \sqrt{169-144}$
$QR = \sqrt{25} = 5\ cm$

Now, according to the question,
$\tan P -\cot R$ = $\frac{RQ}{QP}-\frac{QR}{PQ} = \frac{5}{12} - \frac{5}{12} = 0$

Q3. If $\sin A=\frac{3}{4},$ calculate $\cos A$ and $\tan A$ .

Answer:

Suppose $\Delta$ ABC is a right-angled triangle in which $\angle B = 90°$ and we have $\sin A=\frac{3}{4},$
So,

Let the length of AB be 4 units and the length of BC = 3 units. So, by using Pythagoras' theorem,
$AB = \sqrt{16-9} = \sqrt{7}$ units
Therefore,
$\cos A = \frac{AB}{AC} = \frac{\sqrt{7}}{4}$ and $\tan A = \frac{BC}{AB} = \frac{3}{\sqrt{7}}$

Q4. Given $15 \cot A=8,$ find $\sin A$ and $\sec A$ .

Answer:

We have,
$15 \cot A=8$ $\Rightarrow \cot A =\frac{8}{15}$
It implies that in the triangle ABC, in which $\angle B =90°$. The length of AB is 8 units, and the length of BC = 15 units

Now, by using Pythagoras' theorem,
$AC = \sqrt{64 +225} = \sqrt{289}$
$\Rightarrow AC =17$ units

So, $\sin A = \frac{BC}{AC} = \frac{15}{17}$
and $\sec A = \frac{AC}{AB} = \frac{17}{8}$

Q5. Given $\sec \theta =\frac{13}{12},$ calculate all other trigonometric ratios.

Answer:

We have,
$\sec \theta =\frac{13}{12},$

This means that the hypotenuse of the triangle is 13 units, and the base is 12 units.
Let ABC be a right-angled triangle in which $\angle$ B is 90 and AB is the base, BC is the perpendicular height, and AC is the hypotenuse.

By using Pythagoras' theorem,
$BC = \sqrt{169-144}=\sqrt{25}$
$\Rightarrow BC = 5$ units

Therefore,
$\sin \theta = \frac{BC}{AC}=\frac{5}{13}$
$\cos \theta = \frac{BA}{AC}=\frac{12}{13}$

$\tan \theta = \frac{BC}{AB}=\frac{5}{12}$

$\cot \theta = \frac{BA}{BC}=\frac{12}{5}$

$\sec \theta = \frac{AC}{AB}=\frac{13}{12}$

$\csc \theta = \frac{AC}{BC}=\frac{13}{5}$

Q6. If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A =\angle B$.

Answer:

We have, A and B are two acute angles of triangle ABC and $\cos A =\cos B$.

According to the question, in triangle ABC,
$\cos A =\cos B$
$\frac{AC}{AB}=\frac{BC}{AB}$
$\Rightarrow AC = AB$
Therefore, $\angle$ A = $\angle$ B [angles opposite to equal sides are equal]

Q7. If $\cot \theta =\frac{7}{8},$ evaluate: $(i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ $(ii)\; \cot ^{2}\theta$

Answer:

Given that,
$\cot \theta =\frac{7}{8}$
$\therefore$ perpendicular (AB) = 8 units and Base (BC) = 7 units
Draw a right-angled triangle ABC in which $\angle B =90°$
Now, by using Pythagoras' theorem,
$AC^2 = AB^2+BC^2$
$AC = \sqrt{64 +49} =\sqrt{113}$

So, $\sin \theta = \frac{AB}{AC} = \frac{8}{\sqrt{113}}$
and $\cos \theta = \frac{BC}{AC} = \frac{7}{\sqrt{113}}$

$\Rightarrow \cot \theta =\frac{\cos \theta}{\sin \theta} = \frac{7}{8}$
$(i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$\Rightarrow \frac{(1-\sin^2\theta)}{(1-\cos^2\theta)} = \frac{\cos^2\theta}{\sin^2\theta} = \cot ^2\theta=(\frac{7}{8})^2 = \frac{49}{64}$

$(ii)\; \cot ^{2}\theta$
$=(\frac{7}{8})^2 = \frac{49}{64}$

Q8. If $3\cot A=4,$ check wether $\frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A$ or not.

Answer:

Given that,
$3\cot A=4,$
$\Rightarrow \cot = \frac{4}{3} = \frac{base}{perp.}$
ABC is a right-angled triangle in which $\angle B =90°$ and the length of the base AB is 4 units and the length of the perpendicular is 3 units

By using Pythagoras' theorem, in triangle ABC,
$\\AC^2=AB^2+BC^2\Rightarrow AC = \sqrt{16+9}\Rightarrow AC = \sqrt{25}$
$\Rightarrow AC = 5$ units

So,
$\tan A = \frac{BC}{AB} = \frac{3}{4}$
$\cos A = \frac{AB}{AC} = \frac{4}{5}$
$\sin A = \frac{BC}{AC} = \frac{3}{5}$
$\frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A$
Putting the values of the above trigonometric ratios, we get;
$\Rightarrow \frac{1-\frac{9}{16}}{1+\frac{9}{16}} = \frac{16}{25}-\frac{9}{25}$
$\Rightarrow \frac{7}{25} = \frac{7}{25}$
LHS $=$ RHS

Q9. In triangle $ABC$ , right-angled at $B$ , if $\tan A =\frac{1}{\sqrt{3}},$ find the value of:

$(i) \sin A\: \cos C + \cos A\: \sin C$
$(ii) \cos A\: \cos C + \sin A\: \sin C$

Answer:

Given a triangle ABC, right-angled at B and $\tan A =\frac{1}{\sqrt{3}}$ $\Rightarrow A=30°$

According to question, $\tan A =\frac{1}{\sqrt{3}} = \frac{BC}{AB}$
By using Pythagoras' theorem,
$\\AC^2 = AB^2+BC^2\Rightarrow AC = \sqrt{1+3} =\sqrt{4}$
$\Rightarrow AC = 2$
Now,
$\\\sin A = \frac{BC}{AC} = \frac{1}{2}, \sin C =\frac{AB}{AC} = \frac{\sqrt{3}}{2}, \cos A = \frac{AB}{AC} = \frac{\sqrt{3}}{2}, \cos C = \frac{BC}{AC} = \frac{1}{2}$

Therefore,

$(i) \sin A\: \cos C + \cos A\: \sin C$
$= \frac{1}{2}\times\frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}=\frac{1}{4} +\frac{3}{4} = 1$

$(ii) \cos A\: \cos C + \sin A\: \sin C$
$= \frac{\sqrt{3}}{2}\times \frac{1}{2}+\frac{1}{2}\times\frac{\sqrt{3}}{2}= \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}= \frac{\sqrt{3}}{2}$

Q10. In $\Delta \: PQR$ , right-angled at $Q$ , $PR + QR = 25\: cm$ and $PQ = 5 \: cm$ . Determine the values of $\sin P, \cos P \: and\: \tan P.$

Answer:

We have, PR + QR = 25 cm.............(i)
PQ = 5 cm and $\angle Q =90°$
According to the question,
In triangle $\Delta$ PQR,
By using Pythagoras' theorem,
$PR^2 = PQ^2+QR^2$
$\Rightarrow PQ^2 =PR^2-QR^2$
$\Rightarrow 5^2= (PR-QR)(PR+QR)$
$\Rightarrow 25 = 25(PR-QR)$
$\Rightarrow PR - QR = 1$........(ii)

From equation(i) and equation(ii), we get;
PR = 13 cm and QR = 12 cm.

Therefore,
$\sin P= \frac{QR}{PR}= \frac{12}{13}$
$\cos P = \frac{PQ}{RP} = \frac{5}{13}$
$\tan P = \frac{\sin P}{\cos P} = \frac{12}{5}$

Q11. State whether the following are true or false. Justify your answer.

(i) The value of $\tan A$ is always less than 1.
(ii) $\sec A=\frac{12}{5}$ for some value of angle A.
(iii) $\cos A$ is the abbreviation used for the cosecant of angle A.
(iv) $\cot A$ is the product of cot and A.
(v) $\sin \theta =\frac{4}{3}$ for some angle $\theta .$

Answer:

(i) False,
because $\tan 60 = \sqrt{3}$ , which is greater than 1

(ii) True,
because $\sec A \geq 1$

(iii) False,
Because the $\cos A$ abbreviation is used for cosine A.

(iv) False,
because the term $\cot A$ is a single term, not a product.

(v) False,
because $\sin \theta$ lies between (-1 to +1) [ $-1\leq \sin \theta\leq 1$]

Q1. Evaluate the following :

$(i) \sin 60°\cos 30°+\sin30° \cos 60°$

Answer:
As we know,
the value of $\sin 60° = \frac{\sqrt{3}}{2} = \cos 30°$ , $\sin 30° = \frac{1}{2}=\cos 60°$
$\sin 60°\cos 30°+\sin30° \cos 60°$
$= \frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}+ \frac{1}{2}.\frac{1}{2}\\$
$=\frac{3}{4}+\frac{1}{4}$
$=1$

Q1. Evaluate the following :

$(ii)\: 2\tan ^{2}45°+ 2\cos ^{2}30°- 2\sin ^{2}60°$

Answer:

We know the value of

$\tan 45°= 1$ and

$\cos 30° = \sin 60° = \frac{\sqrt{3}}{2}$
According to the question,

$=2\tan ^{2}45°+ 2\cos ^{2}30°- 2\sin ^{2}60°$
$\\=2(1)^2+ 2(\frac{\sqrt{3}}{2})^2-2(\frac{\sqrt{3}}{2})^2\\=2$

Q1. Evaluate the following :

$(iii)\: \frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$

Answer:
$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$
We know the value of
$\cos 45^{\circ} = \frac{1}{\sqrt{2}}$ , $\sec 30^{\circ}= \frac{2}{\sqrt {3}}$ and $\operatorname{cosec}30^{\circ} =2$ ,
After putting these values
$=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}$
$=\frac{\frac{1}{\sqrt{2}}}{\frac{(2+2\sqrt{3})}{ \sqrt{3}}}$
$\\=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}\times\frac{2\sqrt{2}-2\sqrt{6}}{2\sqrt{2}-2\sqrt{6}}$
$\\=\frac{2\sqrt{6}-2\sqrt{18}}{-16}\\ =2(\frac{\sqrt{6}-3\sqrt{3}}{-16}) = \frac{3\sqrt{3}-\sqrt{6}}{8}$

Q1. Evaluate the following :

$(iv)\: \frac{\sin 30°+\tan 45°-\operatorname{cosec} 60°}{\sec 30°+\cos 60°+\cot 45°}$

Answer:

$\frac{\sin 30°+\tan 45°-\operatorname{cosec} 60°}{\sec 30°+\cos 60°+\cot 45°}$ ..................(i)
It is known that the values of the given trigonometric functions,
$\\\sin 30° = \frac{1}{2}=cos 60°, \tan 45° = 1=\cot 45°, \sec 30° = \frac{2}{\sqrt{3}}=\operatorname{cosec} 60°$
Putting all these values in equation (i), we get;
$\\\Rightarrow \frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}= \frac{3\sqrt{3}-4}{3\sqrt{3}+4}\times\frac{4-3\sqrt{3}}{4-3\sqrt{3}}= \frac{12\sqrt{3}-27-16+12\sqrt{3}}{-11}= \frac{43-24\sqrt{3}}{11}$

Q1. Evaluate the following :

$(v)\frac{5\cos^{2}60°+ 4\sec^{2}30°-\tan^{2}45°}{\sin^{2}30°+\cos^{2}30°}$

Answer:

$\frac{5\cos^{2}60°+ 4\sec^{2}30°-\tan^{2}45°}{\sin^{2}30°+\cos^{2}30°}$ .....................(i)
We know the values of,
$\\\cos 60° = \frac{1}{2}= \sin 30°, \sec 30° = \frac{2}{\sqrt{3}}, \tan 45° = 1, \cos 30° = \frac{\sqrt{3}}{2}$
By substituting all these values in equation(i), we get;

$\\\Rightarrow \frac{5.(\frac{1}{2})^2+4.(\frac{2}{\sqrt{3}}) ^2-1}{(\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}= \frac{\frac{5}{4}-1+\frac{16}{3}}{1}= \frac{\frac{1}{4}+\frac{16}{3}}{1}=\frac{67}{12}$

Q2. Choose the correct option and justify your choice :

$(i)\, \frac{2\: \tan 30°}{1+\tan ^{2}30°}=$

$(A)\: \sin 60°$ $(B)\: \cos 60°$ $(C)\: \tan 60°$ $(D)\: \sin 30°$

Answer:

Put the value of $\tan 30°$ in the given question-
$\frac{2\: \tan 30°}{1+\tan ^{2}30°}=\frac{2\times\frac{1}{\sqrt{3}}}{1+(\frac{1}{\sqrt{3}})^2}=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\frac{\sqrt{3}}{2} = \sin 60°$

The correct option is (A).

Q2. Choose the correct option and justify your choice :

$(ii)\: \frac{1-\tan^{2}45°}{1+ \tan^{2}45°}=$

$(A)\: \tan \: 90°$ $(B)\: 1$ $(C) \: \sin 45°$ $(D) \: 0$

Answer:

$\frac{1-\tan^{2}45°}{1+ \tan^{2}45°}=$
We know that $\tan 45° = 1$
So, $\frac{1-1}{1+1}=0$
The correct option is (D).

Q2. Choose the correct option and justify your choice :

$(iii)\sin \: 2A=2\: \sin A$ is true when $A$ =

$(A)0°$ $(B)\: 30°$ $(C)\: 45°$ $(D)\: 60°$

Answer:

The correct option is (A).
$\sin \: 2A=2\: \sin A$
We know that $\sin 2A = 2\sin A \cos A$
So, $2\sin A \cos A = 2\sin A$
$\\\cos A = 1, A = 0°$

Q2. Choose the correct option and justify your choice :

$(iv)\frac{2\: \tan 30°}{1-tan^{2}\: 30°}=$

$(A)\: \cos 60°$ $(B)\: \sin 60°$ $(C)\: \tan 60°$ $(D)\: \sin 30°$

Answer:

Put the value of $tan 30°={\frac{1}{\sqrt{3}}}$

$\\=\frac{2\: \tan 30°}{1-tan^{2}\: 30°}=\frac{2\times \frac{1}{\sqrt{3}}}{1-(\frac{1}{\sqrt{3}})^2}\\ =\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}\\ =\frac{2}{\sqrt{3}}\times \frac{3}{2}\\ =\sqrt{3}= \tan 60°$

The correct option is (C).

Q3. If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)= \frac{1}{\sqrt{3}};$ $0°<A+B\leq 90°;A> B,$ find $A \: and \: B.$

Answer:

Given that,
$\tan (A+B) = \sqrt{3} =\tan 60°$
So, $A + B = 60^o$ ..........(i)
$\tan (A-B) = \frac{1}{\sqrt{3}} =\tan 30°$
therefore, $A - B = 30^o$ .......(ii)
By solving equations (i) and (ii), we get;

$A = 45^o$ and $B = 15^o$

Q4. State whether the following are true or false. Justify your answer.

$(i) \sin (A + B) = \sin A + \sin B$
$(ii)$ The value of $\sin \theta$ increases as $\theta$ increases.
$(iii)$ The value of $\cos \theta$ increases as $\theta$ increases.
$(iv)\sin \theta =\cos \theta$ for all values of $\theta$ .
$(v) \cot A$ is not defined for $A=0°$

Answer:

(i) False,
Let A = B = $45°$
Then, $\\\sin(45°+45°) = \sin 45°+\sin 45°$
$\sin 90° = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \neq 1$

(ii) True,
Take $\theta = 0°,\ 30°,\ 45°$
when
$\theta$ = 0 then zero(0),
$\theta$ = 30 then value of $\sin \theta$ is $\frac{1}{2}$ = 0.5
$\theta$ = 45 then value of $\sin \theta$ is $\frac{1}{\sqrt2}$ = 0.707

(iii) False,
$\cos 0° = 1,\ \cos 30° = \frac{\sqrt{3}}{2}= 0.87,\ \cos 45° = \frac{1}{\sqrt2}= 0.707$

(iv) False,
Let $\theta = 0°$
$\\\sin 0° = 0$ and $\cos 0°=1$

(v) True,
$\cot 0° = \frac{\cos 0°}{\sin 0°}=\frac{1}{0}$ (not defined).

Q1. Express the trigonometric ratios $\sin A,\sec, and \tan A$ in terms of $\cot A$.

Answer:

We know that $\csc^2A -\cot^2A = 1$
(i)
$\\\Rightarrow \frac{1}{\sin^2A}= 1+\cot^2A\\ \Rightarrow\sin^2A = \frac{1}{1+\cot^2A}\\ \Rightarrow \sin A = \frac{1}{\sqrt{1+\cot^2A}}$

(ii) We know the identity of
$\sec ^2 A-\tan ^2 A=1$
$\Rightarrow \sec^2 A=1+\frac{1}{\cot^2\theta}$
$\Rightarrow \sec A=\frac{\sqrt{\cot^2\theta+1}}{\cot\theta}$

(iii) $\tan A = \frac{1}{\cot A}$

Q2. Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.

Answer:

We know that the identity $\sin^2 A + \cos^2 =1$
$\\\sin^2 A =1- \cos^2 A$
$\sin^2A = 1-\frac{1}{\sec^2A}=\frac{\sec^2A -1}{\sec^2A}$
$\sin A =\sqrt{\frac{\sec^2A -1}{\sec^2A}}=\frac{{}\sqrt{\sec^2A -1}}{\sec A}$
$cosec A =\frac{\sec A}{\sqrt{\sec^2A -1}}$
$\tan A = \frac{\sin A}{\cos A} = \sqrt{\sec^2A -1}$
$\cot A =\frac{1}{ \sqrt{\sec^2A -1}}$
$\cos A=\frac{1}{\sec A}$

Q3. Choose the correct option. Justify your choice.

$(i) 9\sec^{2}A-9\tan^{2}A=$

(A) 1 (B) 9 (C) 8 (D) 0

Answer:

The correct option is (B) = 9

$9\sec^2A-9 \tan ^2A = 9(\sec^2A- \tan ^2A)$ .............(i)

And it is known that sec²A - tan²A = 1

Therefore, equation (i) becomes, $9\times 1 = 9$

Q3. Choose the correct option. Justify your choice.

$(ii)(1+\tan \theta +\sec \theta )(1+\cot \theta -cosec\: \theta )=$

(A) 0 (B) 1 (C) 2 (D) –1

Answer:

The correct option is (C)

$(1+\tan \theta +\sec \theta )(1+\cot \theta -cosec\: \theta )$ .......................(i)

We can write his above equation as;
$=(1+\frac{\sin \theta}{\cos \theta} +\frac{1}{\cos \theta})(1+\frac{\cos\theta}{\sin \theta} -\frac{1}{sin\theta})$
$= \frac{(1+\sin\theta+\cos\theta)}{\cos\theta.\sin\theta}\times (\frac{(\sin\theta+\cos\theta-1}{\sin\theta.\cos\theta})$
$= \frac{(\sin\theta+\cos\theta)^2-1^2}{\sin\theta.\cos\theta}$
$= \frac{\sin^2\theta+\cos^2\theta+2\sin\theta.\\cos\theta-1}{\sin\theta.\cos\theta}$
$= 2\times\frac{\sin\theta.\cos\theta}{\sin\theta.\cos\theta}$
$= 2$

Q3. Choose the correct option. Justify your choice.

$(iii) (\sec A+\tan A)(1-\sin A)=$

$(A)\sec A$ $(B)\sin A$ $(C)cosec A$ $(D) \cos A$

Answer:

The correct option is (D)

$(\sec A+\tan A)(1-\sin A)$
$= ( \frac{1}{\cos A}+\frac{\sin A}{\cos A})(1-\sin A)= \frac{1+\sin A}{\cos A}(1-\sin A)= \frac{1-\sin^2 A}{\cos A}= \cos A$

Q3. Choose the correct option. Justify your choice.

$(iv) \frac{1+\tan ^{2}A}{1+\cot ^{2}A}=$

$(A) \sec ^{2}A$ $(B) -1$ $(C) \cot ^{2}A$ $(D) \tan ^{2}A$

Answer:

The correct option is (D)

$\frac{1+\tan ^{2}A}{1+\cot ^{2}A}$

We know that $\cot A = \frac{1}{\tan A}$

therefore,

$\\\Rightarrow \frac{1+\tan ^{2}A}{1+\frac{1}{\tan ^{2}A}}= \tan^2A\times(\frac{1+\tan^2 A}{1+\tan^2A})= \tan^2A$

Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$(i) (\csc \theta -\cot \theta )^{2}= \frac{1-\cos \theta }{1+\cos \theta }$

Answer:

We need to prove-
$(\csc \theta -\cot \theta )^{2}= \frac{1-\cos \theta }{1+\cos \theta }$

Now, taking LHS,

$(\csc\theta-\cot\theta)^2 = (\frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta})^2$
$\\= (\frac{1-\cos\theta}{\sin\theta})^2\\ =\frac{(1-\cos\theta)(1-\cos\theta)}{\sin^2\theta}\\$
$\\=\frac{(1-\cos\theta)(1-\cos\theta)}{1-\cos^2\theta}\\\\ =\frac{(1-\cos\theta)(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}\\\\ =\frac{1-\cos\theta}{1+\cos\theta}$
LHS = RHS
Hence proved.

Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$(ii) \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}= 2\sec A$

Answer:

We need to prove-

$\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}= 2\sec A$

Taking LHS;
$\\=\frac{\cos^2A+1+\sin^2A+2\sin A}{\cos A(1+\sin A)}\\\\ =\frac{2(1+\sin A)}{\cos A(1+\sin A)}\\\\ =\frac{2}{\cos A} = 2\sec A$

= RHS

Hence proved.

Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$(iii)\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta$

[ Hint: Write the expression in terms of $\sin \theta$ and $\cos\theta$ ]

Answer:

We need to prove-
$\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \:cosec \theta$

Taking LHS;

$\\\Rightarrow \frac{\tan^2 \theta }{\tan \theta-1 }+\frac{1}{\tan\theta(1-\tan \theta)}=\frac{\tan^3\theta-\tan^4\theta+\tan\theta-1}{(\tan\theta-1).\tan\theta.(1-\tan\theta)}= \frac{(\tan^3\theta-1)(1-\tan\theta)}{\tan\theta.(\tan\theta-1)(1-\tan\theta)}\\$
By using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$

$\\\Rightarrow \frac{(\tan\theta -1)(\tan^2\theta+1+\tan\theta)}{\tan\theta(\tan\theta -1)}= \tan\theta+1+\frac{1}{\tan\theta}= 1+\frac{1+\tan^2\theta}{\tan\theta}= 1+\sec^2\theta \times \frac{1}{\tan\theta}= 1+\sec\theta.\csc\theta\\\\ =RHS$
Hence proved.

Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$(iv)\frac{1+\sec A}{\sec A}=\frac{\sin ^{2}A}{1-\cos A}$

[ Hint: Simplify LHS and RHS separately.]

Answer:

We need to prove-
$\frac{1+\sec A}{\sec A}=\frac{\sin ^{2}A}{1-\cos A}$

Taking LHS;

$\\\Rightarrow \frac{1+\sec A}{\sec A}= \frac{(1+\frac{1}{\cos A})}{\sec A}= 1+\cos A$

Taking RHS;
We know that identity $1-\cos^2\theta = \sin^2\theta$

$\\\Rightarrow \frac{\sin ^{2}A}{1-\cos A}= \frac{1-\cos^2 A}{1-\cos A}= \frac{(1-\cos A)(1+\cos A)}{(1-\cos A)}= 1+\cos A$

LHS = RHS

Hence proved.

Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. $(v) \frac{\cos A-\sin A+1}{\cos A+\sin A-1}= \csc A+\cot A$ , using the identity $\csc ^{2}A= 1+\cot ^{2}A$

Answer:

We need to prove -
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}= cosec A+\cot A$

Dividing the numerator and denominator by $\sin A$ , we get;

$\\=\frac{\cot A-1+\csc A}{\cot A +1-\csc A}\\\\= \frac{(\cot A+\csc A)-(\csc^2 A-\cot^2A)}{\cot A +1-\csc A}\\\\= \frac{(\csc A+\cot A)(1-\csc A+\cot A)}{\cot A +1-\csc A}\\\\= \csc A+\cot A\\\\ =RHS$

Hence Proved.

Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$(vi)\sqrt{\frac{1+\sin A}{1-\sin A}}= \sec A+\tan A$

Answer:

We need to prove -
$\sqrt{\frac{1+\sin A}{1-\sin A}}= \sec A+\tan A$
Taking LHS;
By rationalising the denominator, we get;

$\\= \sqrt{\frac{1+\sin A}{1-\sin A}\times \frac{1+\sin A}{1+\sin A}}\\\\ = \sqrt{\frac{(1+\sin A)^2}{1-\sin^2A}}\\\\ =\frac{1+\sin A}{\cos A}\\\\ = \sec A + \tan A\\\\ = RHS$

Hence proved.

Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. $(vii)\frac{\sin \theta -2\sin ^{3}\theta }{2\cos ^{3}\theta -\cos \theta }= \tan \theta$

Answer:

We need to prove -
$\frac{\sin \theta -2\sin ^{3}\theta }{2\cos ^{3}\theta -\cos \theta }= \tan \theta$

Taking LHS;
[we know the identity $\cos2\theta = 2\cos^2\theta-1=\cos^2\theta-\sin^2\theta$ ]

$\\\Rightarrow \frac{\sin \theta(1 -2\sin ^{2}\theta) }{\cos\theta(2\cos ^{2}\theta -1) }= \frac{\sin\theta(\sin^2\theta+\cos^2\theta-2\sin^2\theta)}{\cos\theta.\cos2\theta}= \frac{\sin\theta.\cos2\theta}{\cos\theta.\cos2\theta}= \tan\theta =RHS$

Hence proved.

Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$(viii)(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}= 7+\tan ^{2}A+\cot ^{2}A$

Answer:

Given the equation,
$(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}= 7+\tan ^{2}A+\cot ^{2}A$ ..................(i)

Taking LHS;

$(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}$
$\\\Rightarrow \sin^2 A+\csc^2A +2+\cos^2A+\sec^2A+2= 1+2+2+(1+\cot^2A)+(1+\tan^2A)$
[since $\sin^2\theta +\cos^2\theta = 1, \csc^2\theta-\cot^2\theta =1, \sec^2\theta-\tan^2\theta=1$ ]

$=7+\csc^2A+\tan^2A\\ =RHS$

Hence proved

Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$(ix)\:(cosec A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$

[ Hint: Simplify LHS and RHS separately.]

Answer:

We need to prove-
$(coesc A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$
Taking LHS;
$\\\Rightarrow (cosec A-\frac{1}{\csc A})(\sec A-\frac{1}{\sec A})=\frac{(cosec^2-1)}{cosec A}\times\frac{\sec^2A-1}{\sec A}=\frac{\cot^2A}{cosec A}.\frac{\tan^2A}{\sec A}=\sin A .\cos A$

Taking RHS;

$\\\Rightarrow\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}=\frac{\sin A .\cos A}{\sin^2A+\cos^2A}=\sin A.\cos A$

LHS = RHS

Hence proved.

Q4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$(x) (\frac{1+\tan ^{2}A}{1+\cot ^{2}A})= (\frac{1-\tan A}{1-\cot A})^{2}= \tan ^{2}A$

Answer:

We need to prove,
$(\frac{1+\tan ^{2}A}{1+\cot ^{2}A})= (\frac{1-\tan A}{1-\cot A})^{2}= \tan ^{2}A$

Taking LHS;

$\\\Rightarrow \frac{1+\tan ^{2}A}{1+\cot ^{2}A} = \frac{\sec^2A}{\csc^2A}=\tan^2A$

Taking RHS;

$\\=(\frac{1-\tan A}{1-\cot A})^2\\\\= (\frac{1-\frac{\sin A}{\cos A} }{1-\frac{\cos A}{\sin A}})^2\\\\ = \frac{(\cos A -\sin A)^2(\sin^2A)}{(\sin A-\cos A)^2(\cos^2A)}\\\\ =\tan^2A$

LHS = RHS

Hence proved.

Introduction to Trigonometry Class 10 NCERT solutions: Exercise-wise

Exercise-wise NCERT Solutions of Introduction to Trigonometry Class 10 Maths Chapter 8 are provided in the links below.

• Class 10 Maths NCERT Chapter 8 Exercise 8.1
• Class 10 Maths NCERT Chapter 8 Exercise 8.2
• Class 10 Maths NCERT Chapter 8 Exercise 8.3

Class 10 Maths NCERT Chapter 8: Extra Question

Question:

$(\sin \theta+\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2=$?

Answer:

$(\sin \theta+\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2$
= $(\sin^2 \theta+\operatorname{cosec}^2 \theta+2\sin \theta\operatorname{cosec} \theta)+(\cos^2 \theta+\sec^2 \theta+2\cos \theta\sec \theta)$
= $\sin^2 \theta+\cos^2 \theta+\operatorname{cosec}^2 \theta+2\sin \theta\operatorname{cosec} \theta+\sec^2 \theta+2\cos \theta\sec \theta$
= $1+1+\cot^2 \theta+2+1+\tan^2 \theta+2$
= $7+\cot^2 \theta+\tan^2 \theta$
Hence, the correct answer is $7+\tan^2 \theta+\cot^2 \theta$.

Introduction to Trigonometry Class 10 Chapter 8: Topics

Students will explore the following topics in NCERT Class 10 Maths Chapter 8 Introduction to Trigonometry:

• 8.1 Introduction
• 8.2 Trigonometric Ratios
• 8.3 Trigonometric Ratios of Some Specific Angles
• 8.4 Trigonometric Identities

Introduction to Trigonometry Class 10 NCERT Solutions - Important Formulae

Arc Length in a Circle:

• If an arc of length $l$ in a circle with radius $r$ subtends an angle of $\theta$ radians.

$l = r\theta$ and $θ = \frac{l}{r}$
Conversion Between Radian and Degree Measures:

• Radian Measure = ($\frac{π}{180}$) × Degree Measure
• Degree Measure = ($\frac{180}{π}$) × Radian Measure

Trigonometric Ratios for Right Triangles:

The trigonometric ratios of the angle A in right triangle ABC are defined as follows-

$\begin{aligned} & \text { sine of } \angle A=\frac{\text { side oppositeto angle } A}{\text { hypotenuse }}=\frac{B C}{A C} \\ & \text { cosine of } \angle A=\frac{\text { side adjacent to angle } A}{\text { hypotenuse }}=\frac{A B}{A C} \\ & \text { tangent of } \angle A=\frac{\text { side opposite to angle } A}{\text { side adjacentto angle } A}=\frac{B C}{A B} \\ & \text { cosecant of } \angle A=\frac{\text { hypotenuse }}{\text { side opposite to angle } A}=\frac{A C}{B C} \\ & \text { secant of } \angle A=\frac{\text { hypotenuse }}{\text { side adjacent to angle } A}=\frac{A C}{A B} \\ & \text { cotangent of } \angle A=\frac{\text { side adjacent to angle } A}{\text { side opposite to angle } A}=\frac{A B}{B C}\end{aligned}$

The values of all the trigonometric ratios of 0°, 30°, 45°, 60°, and 90°

Also, in a right triangle with an angle $\theta$

• sin θ = Opposite/Hypotenuse

• cos θ = Adjacent/Hypotenuse

• tan θ = Opposite/Adjacent

• cosec θ = Hypotenuse/Opposite

• sec θ = Hypotenuse/Adjacent

• cot θ = Adjacent/Opposite

Reciprocal Trigonometric Ratios:

• sin θ = 1/(cosec θ)

• cosec θ = 1/(sin θ)

• cos θ = 1/(sec θ)

• sec θ = 1/(cos θ)

• tan θ = 1/(cot θ)

• cot θ = 1/(tan θ)

Trigonometric Ratios of Complementary Angles:

For an angle 'θ', the trigonometric ratios of its complementary angle (90° – θ) are:

• sin (90° – θ) = cos θ

• cos (90° – θ) = sin θ

• tan (90° – θ) = cot θ

• cot (90° – θ) = tan θ

• sec (90° – θ) = cosec θ

• cosec (90° – θ) = sec θ

Trigonometric Identities:

• sin² θ + cos² θ = 1

• sin² θ = 1 – cos² θ

• cos² θ = 1 – sin² θ

• cosec² θ – cot² θ = 1

• cot² θ = cosec² θ – 1

• sec² θ – tan² θ = 1

• tan² θ = sec² θ – 1

NCERT Solutions for Class 10 Maths: Chapter Wise

We at Careers360 compiled all the NCERT class 10 Maths solutions in one place for easy student reference. The following links will allow you to access them.

Also read,

• NCERT Notes Class 10 Maths Chapter 8 Introduction to Trigonometry

• NCERT Exemplar Class 10 Maths Solutions Chapter 8 - Introduction To Trigonometry And Its Equations

NCERT Exemplar Solutions - Subject Wise

Given below are the subject-wise exemplar solutions of class 10 NCERT:

• NCERT Exemplar Class 10th Maths
• NCERT Exemplar Class 10th Science

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 10:

• NCERT Books Class 10 Maths
• NCERT Syllabus Class 10 Maths
• NCERT Books Class 10
• NCERT Syllabus Class 10