NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Chapter 8 explains the relationship between the angles and sides of a right triangle. NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry are required for students taking the Class 10 Board exams. These NCERT solutions for Class 10 Maths chapter 8 are strictly based on the NCERT books for Class 10 Maths. NCERT Class 10 maths solutions chapter 8 provides a thorough explanation of each question in the textbook. Furthermore, NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry provides many tips and tricks for answering questions easily. You can also look through the NCERT solutions for Class 10 for other subjects.

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.1

Q1 In , right-angled at , . Determine :

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,
$\\AC^2 = AB^2 + BC^2\\ AC = \sqrt{AB^2+BC^2}$
Therefore, $AC = \sqrt{576+49}$
$AC = \sqrt{625}$
AC = 25 cm

Now,
(i) $\sin A = P/H = BC/AB = 7/25$
$\cos A = B/H = BA/AC = 24/25$

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

Q2 In Fig. 8.13, find .

Answer:

We have, $\Delta$ PQR is a right-angled triangle, length 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 question,
$\tan P -\cot R$ = $\frac{RQ}{QP}-\frac{QR}{PQ}$
= 5/12 – 5/12 = 0

Q3 If calculate and .

Answer:

Suppose $\Delta$ ABC is a right-angled triangle in which $\angle B = 90^0$ and we have $\sin A=\frac{3}{4},$
So,

Let the length of AB be 4 unit and the length of BC = 3 unit 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 find and .

Answer:

We have,
$15 \: \cot A=8,$ $\Rightarrow \cot A =8/15$
It implies that In the triangle ABC in which $\angle B =90^0$ . The length of AB be 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 calculate all other trigonometric ratios.

Answer:

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

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

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

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 and are acute angles such that , then show that .

Answer:

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

According to question, In triangle ABC,
$\cos A =\cos B$

$\frac{AC}{AB}=\frac{BC}{AB}$
$\Rightarrow AC = AB$
Therefore, $\angle$ A = $\angle$ B [angle opposite to equal sides are equal]

Q7 If evaluate:

Answer:

Given that,
$\cot \theta =\frac{7}{8}$
$\therefore$ perpendicular (AB) = 8 units and Base (AB) = 7 units
Draw a right-angled triangle ABC in which $\angle B =90^0$
Now, By using Pythagoras theorem,

$AC = \sqrt{64 +49} =\sqrt{113}$

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

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

Q8 If check wether 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 $\angle B =90^0$ and the length of the base AB is 4 units and length of perpendicular is 3 units

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

So,
$\tan A = \frac{BC}{AB} = \frac{3}{4}$
$\cos A = \frac{AB}{AC} = \frac{4}{5}$
$\sin A = \frac{BC}{AC} = \frac{3}{5}$
$\frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A$
Put the values of above trigonometric ratios, we get;
$\Rightarrow \frac{1-9/4}{1+9/4} = \frac{16}{25}-\frac{9}{25}$
$\Rightarrow -\frac{5}{13} \neq \frac{7}{25}$
LHS $\neq$ RHS

Q9 In triangle , right-angled at , if find the value of:

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,
$\\AC^2 = AB^2+BC^2\\ AC = \sqrt{1+3} =\sqrt{4}$
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,

$(i) \sin A\: \cos C + \cos A\: \sin C$
$\\\Rightarrow \frac{1}{2}\times\frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\\ \Rightarrow1/4 +3/4\\ \Rightarrow4/4 = 1$

\\\Rightarrow \frac{\sqrt{3}}{2}\times \frac{1}{2}+\frac{1}{2}\times\frac{\sqrt{3}}{2}\\ \Rightarrow \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\\ \Rightarrow \frac{\sqrt{3}}{2}

Q10 In , right-angled at , and . Determine the values of

Answer:

We have, PR + QR = 25 cm………….(i)
PQ = 5 cm
and $\angle Q =90^0$
According to question,
In triangle $\Delta$ PQR,
By using Pythagoras theorem,
$\\PR^2 = PQ^2+QR^2\\ PQ^2 =PR^2-QR^2 \\ 5^2= (PR-QR)(PR+QR)\\ 25 = 25(PR-QR) \\$
PR – QR = 1……..(ii)

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

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

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 $\cos A$ 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) [ $-1\leq \sin \theta\leq 1$ ]

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.2

Q1 Evaluate the following :

(i) \sin 60^{o}\cos 30^{o}+\sin30^{o} \cos 60^{o}

Answer:

$\sin 60^{o}\cos 30^{o}+\sin30^{o} \cos 60^{o}$
As we know,
the value of $\sin 60^0 = \sqrt{3}/2 = \cos 30^0$ , $\sin 30^0 = 1/2=\cos 60^0$
$\\\Rightarrow \frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}+ \frac{1}{2}.\frac{1}{2}\\$
$=\frac{3}{4}+\frac{1}{4}$

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 question,

=2\tan ^{2}45^{o}+ 2\cos ^{2}30^{o}- 2\sin ^{2}60^{o}

\\=2(1)^2+ (\frac{\sqrt{3}}{2})^2-(\frac{\sqrt{3}}{2})\\=2

Q1 Evaluate the following :

(iii)\: \frac{\cos 45^{o}}{\sec 30^{o}+\csc 30^{o}}

Answer:

$\frac{\cos 45^{o}}{\sec 30^{o}+\csc 30^{o}}$
we know the value of

$\cos 45 = 1/\sqrt{2}$ , $\sec 30^0 = 2/\sqrt {3}$ and $cosec \:30 =2$ ,

After putting these values
$=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}$
$=\frac{1/\sqrt{3}}{(2+2\sqrt{3})/ \sqrt{3}}$
$\\=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}\times\frac{2\sqrt{2}-2\sqrt{6}}{2\sqrt{2}-2\sqrt{6}}$
$\\=\frac{2\sqrt{6}-2\sqrt{18}}{-16}\\ =2\frac{\sqrt{6}-3\sqrt{3}}{-16} = \frac{3\sqrt{3}-\sqrt{6}}{8}$

Q1 Evaluate the following :

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

Answer:

$\frac{\sin 30^{o}+\tan 45^{o}-cosec 60^{o}}{\sec 30^{o}+\cos 60^{o}+\cot 45^{o}}$ ………………(i)
It is known that the values of the given trigonometric functions,
$\\\sin 30^0 = 1/2=cos 60^0\\ \tan 45^0 = 1=\cot 45^0\\ \sec 30^0 = 2/\sqrt{3}=cosec 60^0\\$
Put all these values in equation (i), we get;
$\\\Rightarrow \frac{1/2+1-2/\sqrt{3}}{2/\sqrt{3}+1/2+1}\\ \Rightarrow\frac{3/2-2/\sqrt{3}}{3/2+2/\sqrt{3}}\\ \Rightarrow \frac{3\sqrt{3}-4}{3\sqrt{3}+4}\times\frac{4-3\sqrt{3}}{4-3\sqrt{3}}\\ \Rightarrow \frac{12\sqrt{3}-27-16+12\sqrt{3}}{-11}\\ \Rightarrow \frac{43-24\sqrt{3}}{11}$

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-
$\\\cos 60^0 = 1/2= \sin 30^0\\ \sec 30^0 = 2/\sqrt{3}\\ \tan 45^0 = 1\\ \cos 30^0 = \sqrt{3}/2$
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}\\ \Rightarrow \frac{5/4-1+16/3}{1}\\ \Rightarrow \frac{1/4+16/3}{1}\\ \Rightarrow\frac{67}{12}

Q2 Choose the correct option and justify your choice :

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

Answer:

Put the value of tan 30 in the given question-
$\frac{2\: \tan 30^{o}}{1+\tan ^{2}30^{o}}=\frac{2\times1/\sqrt{3}}{1+(1/\sqrt{3})^2}=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\sqrt{3}/2 = \sin 60^0$

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}}=

Answer:

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

Q2 Choose the correct option and justify your choice :

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

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-tan^{2}\: 30^{o}}=

Answer:

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

\\=\frac{2\: \tan 30^{o}}{1-tan^{2}\: 30^{o}}=\frac{2\times 1/\sqrt{3}}{1-(\frac{1}{\sqrt{3}})^2}\\ =\frac{2/\sqrt{3}}{1-1/3}\\ =\frac{2}{\sqrt{3}}\times \frac{3}{2}\\ =\sqrt{3}= \tan 60^0

The correct option is (C)

Q3 If and find

Answer:

Given that,
$\tan (A+B) = \sqrt{3} =\tan 60^0$
So, ……….(i)
$\tan (A-B) = 1/\sqrt{3} =\tan 30^0$
therefore, …….(ii)
By solving the equation (i) and (ii) we get;

and

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, $\\\sin(45^0+45^0) = \sin 45^0+\sin 45^0\\ \sin 90^ = 1/\sqrt{2}+q/\sqrt{2}\\ 1 \neq \sqrt{2}$

(ii) True,
Take $\theta = 0^0,\ 30^0,\ 45^0$
whent
$\theta$ = 0 then zero(0),
$\theta$ = 30 then value of $\sin \theta$ is 1/2 = 0.5
$\theta$ = 45 then value of $\sin \theta$ is 0.707

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

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

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

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.3

Q1 Evaluate :

Answer:

We can write the above equation as;
$=\frac{\sin (90^0-72^0)}{\cos 72^0}$
By using the identity of $\sin (90^o-\theta) = \cos \theta$
Therefore, $\frac{\cos 72^0}{\cos 72^0} = 1$

So, the answer is 1.

Q1 Evaluate :

Answer:

$\frac{\tan 26^{o}}{\cot 64^{o}}$
The above equation can be written as ;

$\tan (90^o-64^o)/\cot 64^o$ ………(i)
It is known that, $\tan (90^o-\theta) = \cot \theta$
Therefore, equation (i) becomes,
$\cot64^o/\cot 64^o = 1$

So, the answer is 1.

Q1 Evaluate :

Answer:

$\cos 48^{o}-\sin 42^{o}$
The above equation can be written as ;
$\cos (90^o-42^{o})-\sin 42^{o}$ ………………..(i)
It is known that $\cos (90^o-\theta) = \sin \theta$
Therefore, equation (i) becomes,
$\sin42^{o}-\sin 42^{o} = 0$

So, the answer is 0.

Q1 Evaluate :

Answer:

This equation can be written as;
$cosec 31^o - \sec(90^o-31^o)$ ……………..(i)
We know that $\sec(90^o-\theta) = cosec \theta$

Therefore, equation (i) becomes;
$cosec 31^o - cosec\ 31^o$ = 0

So, the answer is 0.

Q2 Show that :

(i) \tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1

Answer:

$\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1$
Taking Left Hand Side (LHS)
= $\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}$
$\Rightarrow \tan 48^{o}\tan 23^{o}\tan (90^o-48^{o})\tan (90^o-23^{o})$
$\Rightarrow \tan 48^{o}\tan 23^{o}\cot 48^{o}\cot23^{o}$ [it is known that $\tan (90^0-\theta = \cot\theta)$ and $\cot\theta\times \tan \theta =1$

Hence proved.

Q2 Show that :

(ii) \cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0

Answer:

\cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0

Taking Left Hand Side (LHS)
= $\cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}$
= $\cos 38^{o}\cos (90^o-38^{o})-\sin 38^{o}\sin (90^o-38^{o})$
= $\cos 38^{o}\sin38^{o}-\sin 38^{o}\cos 38^{o}$ [it is known that $\sin(90^0-\theta) =\cos \theta$ and $\cos(90^0-\theta) =\sin \theta$ ]
= 0

Q3 If , where is an acute angle, find the value of .

Answer:

We have,
$\tan$ 2A = $\cot$ (A – $\18^{0}$ )
we know that, $\\\cot (90^0-\theta) = \tan\theta$
$\\\Rightarrow \cot (90^0-2A) = \cot (A-18^0)\\\Rightarrow 90^0-2A = A - 18^0\\\Rightarrow 3A = 108\\\Rightarrow A = 108/3 = 36^0$

Q4 If , prove that .

Answer:

We have,
$\tan A= \cot B$
and we know that $\tan (90^0 - \theta)= \cot \theta$
therefore,
$\tan A= \tan(90^0- B)$
A = 90 – B
A + B = 90
Hence proved.

Q5 If , where is an acute angle, find the value of .

Answer:

We have,
$\sec 4A= cosec (A-20^{o})$ , Here 4A is an acute angle
According to question,
We know that $cosec(90^0-\theta)= \sec \theta$
$cosec(90^0-4A)= cosec (A-20^{o})$

\\\Rightarrow 90 - 4A = A - 20\\\\ \Rightarrow 5A=110\\\\ \Rightarrow A=\frac{110}{5}\\\:\\ \Rightarrow A=22^o

Q6 If and are interior angles of a triangle , then show that

Answer:

Given that,
A, B and C are interior angles of $\Delta ABC$
To prove – $\sin (\frac{B+C}{2})= \cos \frac{A}{2}$

Now,
In triangle $\Delta ABC$ ,
A + B + C = 180^0
$\Rightarrow B + C = 180 - A$
$\Rightarrow B + C/2 = 90^0 - A/2$
$\sin \frac{B+C}{2}=\sin (90^0-A/2)$
$\sin \frac{B+C}{2}=\cos A/2$
Hence proved.

Q7 Express in terms of trigonometric ratios of angles between and .

Answer:

By using the identity of $\sin\theta$ and $\cos\theta$
$sin 67^{o}+\cos 75^{o}$
We know that,
$\sin(90-\theta) =\cos \theta$ and $\cos(90-\theta) =\sin \theta$
the above equation can be written as;
$=\sin (90^0-23^0)+\cos(90^0-15^0)$
$=\sin (15^0)+\cos(23^0)$

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Excercise: 8.4

Q1 Express the trigonometric ratios and in terms of .

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

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

Q2 Write all the other trigonometric ratios of in terms of .

Answer:

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

cosec A =\frac{\sec A}{\sqrt{\sec^2A -1}}

\tan A = \frac{\sin A}{\cos A} = \sqrt{\sec^2A -1}

Q3 Evaluate :

(i)\frac{\sin ^{2}63^{o}+\sin ^{2}27^{o}}{\cos ^{2}17^{o}+\cos ^{2}73^{o}}

Answer:

$\frac{\sin ^{2}63^{o}+\sin ^{2}27^{o}}{\cos ^{2}17^{o}+\cos ^{2}73^{o}}$ ………………..(i)

The above equation can be written as;

$\\=\frac{\sin ^{2}63^{o}+\sin ^{2}(90^0-63^{o})}{\cos ^{2}(90^0-73^{o})+\cos ^{2}73^{o}}\\\\ =\frac{\sin ^{2}63^{o}+\cos ^{2}63^{o}}{\sin ^{2}73^{o}+\cos ^{2}73^{o}}\\\\ = 1$
(Since $\sin^2\theta +\cos^2\theta = 1$ )

Q3 Evaluate :

(ii)\sin 25^{o}\cos 65^{o}+\cos 25^{o}\sin 65^{o}

Answer:

\sin 25^{o}\cos 65^{o}+\cos 25^{o}\sin 65^{o}

We know that

Therefore,

\\\sin 25^{o}\cos (90^0-25^{o})+\cos 25^{o}\sin (90^0-25^{o})\\ \sin 25^0.\sin 25^0 + \cos 25^0.\cos 25^0\\ sin^2 25^0+\cos ^225^0\\ 1

Q4 Choose the correct option. Justify your choice.

(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, $9\times 1 = 9$

Q4 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;
$\\=(1+\sin \theta/\cos \theta +1/\cos \theta )(1+\cos\theta/\sin \theta -1/sin\theta )\\\\= \frac{(1+\sin\theta+\cos\theta)}{\cos\theta.\sin\theta}\times (\frac{(\sin\theta+\cos\theta-1}{\sin\theta.\cos\theta})\\\\= \frac{(\sin\theta+\cos\theta)^2-1^2}{\sin\theta.\cos\theta}\\\\= \frac{\sin^2\theta+\cos^2\theta+2\sin\theta.\\cos\theta-1}{\sin\theta.\cos\theta}\\\\= 2\times\frac{\sin\theta.\cos\theta}{\sin\theta.\cos\theta}$
= 2

Q4 Choose the correct option. Justify your choice.

Answer:

The correct option is (D)

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

Q4 Choose the correct option. Justify your choice.

Answer:

The correct option is (D)

$\frac{1+\tan ^{2}A}{1+\cot ^{2}A}$ ……………………..eq (i)

The above equation can be written as;

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

therefore,

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

Q5 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.

Q5 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;
$\\=\frac{\cos^2A+1+\sin^2A+2\sin A}{\cos A(1+\sin A)}\\\\ =\frac{2(1+\sin A)}{\cos A(1+\sin A)}\\\\ =2/\cos A = 2\sec A$

= RHS

Hence proved.

Q5 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;

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

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

Hence proved.

Q5 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}\\ \Rightarrow (1+\frac{1}{\cos A})/\sec A\\ \Rightarrow 1+\cos A

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

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

LHS = RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. , using the identity

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.

Q5 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.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

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

Hence proved.

Q5 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 equation,
$(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}= 7+\tan ^{2}A+\cot ^{2}A$ ………………(i)

Taking LHS;

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

Hence proved

Q5 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;
$\\\Rightarrow (cosec A-\frac{1}{\csc A})(\sec A-\frac{1}{\sec A})\\\\ \Rightarrow\frac{(cosec^2-1)}{cosec A}\times\frac{\sec^2A-1}{\sec A}\\\\ \Rightarrow\frac{\cot^2A}{cosec A}.\frac{\tan^2A}{\sec A}\\\\ \Rightarrow\sin A .\cos A$

Taking RHS;

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

LHS = RHS

Hence proved.

Q5 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-\sin A/\cos A }{1-\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.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.1

Q1 In , right-angled at , . Determine :

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NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.2

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Q3 If and find

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NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.3

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Excercise: 8.4

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