NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Have you ever wondered how engineers design huge buildings, how sailors and pilots navigate through their journey, or how shadows change their length throughout the day? All of these answers can be found in Trigonometry, a fascinating branch of mathematics. As per the latest NCERT syllabus of class 10, this chapter contains the basic concepts of trigonometry, like trigonometric ratios, trigonometric identities, trigonometric ratios of some specific angles, and trigonometric ratios of complementary angles. Understanding these concepts will make students more efficient in solving problems involving height, distance, and angles. This chapter will also build a strong foundation for advanced trigonometric concepts, which have many practical real-life applications like construction, navigation, architecture, etc.

The NCERT solutions of class 10 Maths Chapter 8, Introduction to Trigonometry, are designed by our experienced subject experts to offer clear and step-by-step solutions for the exercise problems and help students to prepare well for exams and to gain knowledge about all the natural processes happening around them through a series of solved questions. It covers questions from all the topics and will help you improve your speed and accuracy.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Download PDF

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

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

Trigonometric Ratios for Right Triangles:

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

$\begin{array}{rl}& \text{ sine of }\mathrm{\angle }A=\frac{\text{ side oppositeto angle }A}{\text{ hypotenuse }}=\frac{BC}{AC}\\ & \text{ cosine of }\mathrm{\angle }A=\frac{\text{ side adjacent to angle }A}{\text{ hypotenuse }}=\frac{AB}{AC}\\ & \text{ tangent of }\mathrm{\angle }A=\frac{\text{ side opposite to angle }A}{\text{ side adjacentto angle }A}=\frac{BC}{AB}\\ & \text{ cosecant of }\mathrm{\angle }A=\frac{\text{ hypotenuse }}{\text{ side opposite to angle }A}=\frac{AC}{BC}\\ & \text{ secant of }\mathrm{\angle }A=\frac{\text{ hypotenuse }}{\text{ side adjacent to angle }A}=\frac{AC}{AB}\\ & \text{ cotangent of }\mathrm{\angle }A=\frac{\text{ side adjacent to angle }A}{\text{ side opposite to angle }A}=\frac{AB}{BC}\end{array}$

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

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 8 Introduction to Trigonometry (Exercises)

NCERT introduction to trigonometry class 10 questions and answers
Exercise: 8.1, Page number: 121, Total questions: 11

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

Answer:

We have,
In $\mathrm{\Delta }ABC$ , $\mathrm{\angle }$B = 90, and the length of the base (AB) = 24 cm and length of perpendicular (BC) = 7 cm
So, by using Pythagoras' theorem,
$A{C}^2=A{B}^2+B{C}^2$
⇒ $AC=\sqrt{A{B}^2+B{C}^2}$
⇒ $AC=\sqrt{576+49}$
⇒ $AC=\sqrt{625}$
⇒ $AC=25$ 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 $\mathrm{\angle }$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 $\mathrm{\Delta }$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,
$QR=\sqrt{{13}^2-{12}^2}$
$QR=\sqrt{169-144}$
$QR=\sqrt{25}=5cm$

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 $\mathrm{\Delta }$ ABC is a right-angled triangle in which $\mathrm{\angle }B={90}^0$ 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 $\mathrm{\angle }B={90}^0$. 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 $\mathrm{\angle }$ 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 $\mathrm{\angle }A$ and $\mathrm{\angle }B$ are acute angles such that $\cos A=\cos B$, then show that $\mathrm{\angle }A=\mathrm{\angle }B$.

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, $\mathrm{\angle }$ A = $\mathrm{\angle }$ 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 $\mathrm{\angle }B={90}^0$
Now, by using Pythagoras' theorem,
$A{C}^2=A{B}^2+B{C}^2$
$AC=\sqrt{64+49}=\sqrt{113}$

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 $\mathrm{\angle }B={90}^0$ 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,
$A{C}^2=A{B}^2+B{C}^2\Rightarrow AC=\sqrt{16+9}\Rightarrow AC=\sqrt{25}$
$\Rightarrow AC=5$ 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 $\tan A=\frac{1}{\sqrt{3}}$ $\Rightarrow A={30}^0$

According to question, $\tan A=\frac{1}{\sqrt{3}}=\frac{BC}{AB}$
By using Pythagoras' theorem,
$A{C}^2=A{B}^2+B{C}^2\Rightarrow AC=\sqrt{1+3}=\sqrt{4}$
$\Rightarrow AC=2$
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 $\mathrm{\Delta }PQR$ , right-angled at $Q$ , $PR+QR=25cm$ and $PQ=5cm$ . Determine the values of $\sin P,\cos Pand\tan P.$

Answer:

We have, PR + QR = 25 cm.............(i)
PQ = 5 cm and $\mathrm{\angle }Q={90}^0$
According to the question,
In triangle $\mathrm{\Delta }$ PQR,
By using Pythagoras' theorem,
$P{R}^2=P{Q}^2+Q{R}^2$
$\Rightarrow P{Q}^2=P{R}^2-Q{R}^2$
$\Rightarrow 5^2=(PR-QR)(PR+QR)$
$\Rightarrow 25=25(PR-QR)$
$\Rightarrow PR-QR=1$........(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 $\sec A\ge 1$

(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) [ $-1\le \sin \theta \le 1$]

NCERT introduction to trigonometry class 10 questions and answers
Exercise: 8.2, Page number: 127, Total questions: 4

Q1. Evaluate the following :

$(i)\sin {60}^o\cos {30}^o+\sin {30}^o\cos {60}^o$

Answer:
As we know,
the value of $\sin {60}^0=\frac{\sqrt{3}}{2}=\cos {30}^0$ , $\sin {30}^0=\frac{1}{2}=\cos {60}^0$
$\sin {60}^o\cos {30}^o+\sin {30}^o\cos {60}^o$
$=\frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}+\frac{1}{2}.\frac{1}{2}$
$=\frac{3}{4}+\frac{1}{4}$
$=1$

Q1. Evaluate the following :

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

Answer:

We know the value of

$\tan {45}^0=1$ and

$\cos {30}^0=\sin {60}^0=\frac{\sqrt{3}}{2}$
According to the question,

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

Q1. Evaluate the following :

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

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

Q1. Evaluate the following :

$(iv)\frac{\sin {30}^o+\tan {45}^o-\mathrm{cosec}{60}^o}{\sec {30}^o+\cos {60}^o+\cot {45}^o}$

Answer:

$\frac{\sin {30}^o+\tan {45}^o-\mathrm{cosec}{60}^o}{\sec {30}^o+\cos {60}^o+\cot {45}^o}$ ..................(i)
It is known that the values of the given trigonometric functions,
$\sin {30}^0=\frac{1}{2}=cos{60}^0,\tan {45}^0=1=\cot {45}^0,\sec {30}^0=\frac{2}{\sqrt{3}}=\mathrm{cosec}{60}^0$
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 :

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

Answer:

$\frac{5{\cos }^2{60}^o+4{\sec }^2{30}^o-{\tan }^2{45}^o}{{\sin }^2{30}^o+{\cos }^2{30}^o}$ .....................(i)
We know the values of,
$\cos {60}^0=\frac{1}{2}=\sin {30}^0,\sec {30}^0=\frac{2}{\sqrt{3}},\tan {45}^0=1,\cos {30}^0=\frac{\sqrt{3}}{2}$
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 :

$(i)\frac{2\tan {30}^o}{1+{\tan }^2{30}^o}=$

$(A)\sin {60}^o$ $(B)\cos {60}^o$ $(C)\tan {60}^o$ $(D)\sin {30}^o$

Answer:

Put the value of $\tan {30}^0$ in the given question-
$\frac{2\tan {30}^o}{1+{\tan }^2{30}^o}=\frac{2\times \frac{1}{\sqrt{3}}}{1+(\frac{1}{\sqrt{3}}{)}^2}=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\frac{\sqrt{3}}{2}=\sin {60}^0$

The correct option is (A).

Q2. Choose the correct option and justify your choice :

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

$(A)\tan {90}^o$ $(B)1$ $(C)\sin {45}^o$ $(D)0$

Answer:

$\frac{1-{\tan }^2{45}^o}{1+{\tan }^2{45}^o}=$
We know that $\tan {45}^0=1$
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 $A$ =

$(A)0^o$ $(B){30}^o$ $(C){45}^o$ $(D){60}^o$

Answer:

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

Q2. Choose the correct option and justify your choice :

$(iv)\frac{2\tan {30}^o}{1-ta{n}^2{30}^o}=$

$(A)\cos {60}^o$ $(B)\sin {60}^o$ $(C)\tan {60}^o$ $(D)\sin {30}^o$

Answer:

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

$$\begin{gathered} =\frac{2\tan {30}^o}{1-ta{n}^2{30}^o}=\frac{2\times \frac{1}{\sqrt{3}}}{1-(\frac{1}{\sqrt{3}}{)}^2} \\ =\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}} \\ =\frac{2}{\sqrt{3}}\times \frac{3}{2} \\ =\sqrt{3}=\tan {60}^0 \end{gathered}$$

The correct option is (C).

Q3. If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=\frac{1}{\sqrt{3}};$ $0^o<A+B\le {90}^o;A>B,$ find $AandB.$

Answer:

Given that,
$\tan (A+B)=\sqrt{3}=\tan {60}^0$
So, $A+B={60}^o$ ..........(i)
$\tan (A-B)=\frac{1}{\sqrt{3}}=\tan {30}^0$
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 $A=0^o$

Answer:

(i) False,
Let A = B = ${45}^0$
Then, $\sin ({45}^0+{45}^0)=\sin {45}^0+\sin {45}^0$
$\sin {90}^0=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\ne 1$

(ii) True,
Take $\theta =0^0,{30}^0,{45}^0$
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 $\frac{1}{\sqrt{2}}$ = 0.707

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

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

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

NCERT introduction to trigonometry class 10 questions and answers

Exercise: 8.3, Page number: 131, Total questions: 4

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

Answer:

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

(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 $\mathrm{\angle }A$ in terms of $\sec A$.

Answer:

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

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

$9{\sec }^{2}A-9{\tan }^{2}A=9({\sec }^{2}A-{\tan }^{2}A)$ .............(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;
$=(1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta })(1+\frac{\cos \theta }{\sin \theta }-\frac{1}{sin\theta })$
$=\frac{(1+\sin \theta +\cos \theta )}{\cos \theta .\sin \theta }\times (\frac{(\sin \theta +\cos \theta -1}{\sin \theta .\cos \theta })$
$=\frac{(\sin \theta +\cos \theta {)}^2-1^2}{\sin \theta .\cos \theta }$
$$\begin{gathered} =\frac{{\sin }^2\theta +{\cos }^2\theta +2\sin \theta . \\ cos\theta -1}{\sin \theta .\cos \theta } \end{gathered}$$
$=2\times \frac{\sin \theta .\cos \theta }{\sin \theta .\cos \theta }$
$=2$

Q3. Choose the correct option. Justify your choice.

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

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

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,

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

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,

$(\csc \theta -\cot \theta {)}^2=(\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }{)}^2$
$$\begin{gathered} =(\frac{1-\cos \theta }{\sin \theta }{)}^2 \\ =\frac{(1-\cos \theta )(1-\cos \theta )}{{\sin }^2\theta } \end{gathered}$$
$$\begin{gathered} =\frac{(1-\cos \theta )(1-\cos \theta )}{1-{\cos }^2\theta } \\ =\frac{(1-\cos \theta )(1-\cos \theta )}{(1-\cos \theta )(1+\cos \theta )} \\ =\frac{1-\cos \theta }{1+\cos \theta } \end{gathered}$$
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;
$$\begin{gathered} =\frac{{\cos }^{2}A+1+{\sin }^{2}A+2\sin A}{\cos A(1+\sin A)} \\ =\frac{2(1+\sin A)}{\cos A(1+\sin A)} \\ =\frac{2}{\cos A}=2\sec A \end{gathered}$$

= 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 $a^3-b^3=(a-b)(a^2+ab+b^2)$

$$\begin{gathered} \Rightarrow \frac{(\tan \theta -1)({\tan }^2\theta +1+\tan \theta )}{\tan \theta (\tan \theta -1)}=\tan \theta +1+\frac{1}{\tan \theta }=1+\frac{1+{\tan }^2\theta }{\tan \theta }=1+{\sec }^2\theta \times \frac{1}{\tan \theta }=1+\sec \theta .\csc \theta \\ =RHS \end{gathered}$$
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 -
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosecA+\cot A$

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

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

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;

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

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 $\cos 2\theta =2{\cos }^2\theta -1={\cos }^2\theta -{\sin }^2\theta$ ]

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

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;

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

$$\begin{gathered} =7+{\csc }^{2}A+{\tan }^{2}A \\ =RHS \end{gathered}$$

Hence proved

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

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

[ Hint: Simplify LHS and RHS separately.]

Answer:

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

Taking RHS;

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

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;

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

Taking RHS;

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

LHS = RHS

Hence proved.

Trigonometry Class 10 Solutions - Exercise Wise

Here are the exercise-wise links for the NCERT class 10 chapter 8, Introduction to Trigonometry:

• Introduction To Trigonometry Class 10 Exercise-8.1
• Introduction To Trigonometry Class 10 Exercise-8.2
• Introduction To Trigonometry Class 10 Exercise-8.3

NCERT Solutions for Class 10 Maths: Chapter Wise

Here, students can find chapter-wise solutions.

NCERT Solutions of Class 10 - Subject Wise

Here are the subject-wise links for the NCERT solutions of class 10:

• NCERT solutions for class 10 maths
• NCERT solutions for class 10 science

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