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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. From the latest NCERT syllabus for class 10, the chapter introduction to trigonometry 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.
This article on NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry offers clear and step-by-step solutions for the exercise problems in the NCERT Class 10 Maths Book. Students who are in need of introduction to trigonometry class 10 solutions will find this article very useful. It covers all the important class 10 maths chapter 8 question answers of introduction to trigonometry. These class 10 Introduction to Trigonometry ncert solutions are made by the Subject Matter Experts according to the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. NCERT solutions for class 10 maths and NCERT solutions for other subjects and classes can be downloaded from the NCERT Solutions.
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:
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° are-
$\angle A$ | 0° | 30° | 45° | 60° | 90° |
Sin A | 0 | $\frac{1}{2}$ | $\frac{1}{\sqrt{2}}$ | $\frac{\sqrt{3}}{2}$ | 1 |
Cos A | 1 | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{2}}$ | $\frac{1}{2}$ | 0 |
Tan A | 0 | $\frac{1}{\sqrt{3}}$ | 1 | $\sqrt{3}$ | Not Defined |
Cosec A | Not Defined | 2 | $\sqrt{2}$ | $\frac{2}{\sqrt{3}}$ | 1 |
Sec A | 1 | $\frac{2}{\sqrt{3}}$ | $\sqrt{2}$ | 2 | Not Defined |
Cot A | Not Defined | $\sqrt{3}$ | 1 | $\frac{1}{\sqrt{3}}$ | 0 |
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 introduction to trigonometry class 10 questions and answers
Exercise: 8.1, Page number: 121, Total questions: 11
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 $\angle B = 90^0$ 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 $\angle B =90^0$. 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 $\angle$ 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,
$\sin \theta = \frac{BC}{AC}=\frac{5}{13}$
$\cos \theta = \frac{BA}{AC}=\frac{12}{13}$
$\tan \theta = \frac{BC}{AB}=\frac{5}{12}$
$\cot \theta = \frac{BA}{BC}=\frac{12}{5}$
$\sec \theta = \frac{AC}{AB}=\frac{13}{12}$
$\csc \theta = \frac{AC}{BC}=\frac{13}{5}$
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 $\angle B =90^0$
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 $\angle B =90^0$ 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,
$\\AC^2=AB^2+BC^2\Rightarrow AC = \sqrt{16+9}\Rightarrow AC = \sqrt{25}$
$\Rightarrow AC = 5$ 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 $\tan A =\frac{1}{\sqrt{3}}$ $\Rightarrow A=30^0$
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,
$(i) \sin A\: \cos C + \cos A\: \sin C$
$= \frac{1}{2}\times\frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}=\frac{1}{4} +\frac{3}{4} = 1$
$(ii) \cos A\: \cos C + \sin A\: \sin C$
$= \frac{\sqrt{3}}{2}\times \frac{1}{2}+\frac{1}{2}\times\frac{\sqrt{3}}{2}= \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}= \frac{\sqrt{3}}{2}$
Answer:
We have, PR + QR = 25 cm.............(i)
PQ = 5 cm and $\angle Q =90^0$
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) [ $-1\leq \sin \theta\leq 1$]
NCERT introduction to trigonometry class 10 questions and answers
Exercise: 8.2, Page number: 127, Total questions: 4
$(i) \sin 60^{o}\cos 30^{o}+\sin30^{o} \cos 60^{o}$
Answer:
As we know,
the value of $\sin 60^0 = \frac{\sqrt{3}}{2} = \cos 30^0$ , $\sin 30^0 = \frac{1}{2}=\cos 60^0$
$\sin 60^{o}\cos 30^{o}+\sin30^{o} \cos 60^{o}$
$= \frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}+ \frac{1}{2}.\frac{1}{2}\\$
$=\frac{3}{4}+\frac{1}{4}$
$=1$
$(ii)\: 2\tan ^{2}45^{o}+ 2\cos ^{2}30^{o}- 2\sin ^{2}60^{o}$
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,
$=2\tan ^{2}45^{o}+ 2\cos ^{2}30^{o}- 2\sin ^{2}60^{o}$
$\\=2(1)^2+ 2(\frac{\sqrt{3}}{2})^2-2(\frac{\sqrt{3}}{2})^2\\=2$
$(iii)\: \frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$
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
$=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}$
$=\frac{\frac{1}{\sqrt{2}}}{\frac{(2+2\sqrt{3})}{ \sqrt{3}}}$
$\\=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}\times\frac{2\sqrt{2}-2\sqrt{6}}{2\sqrt{2}-2\sqrt{6}}$
$\\=\frac{2\sqrt{6}-2\sqrt{18}}{-16}\\ =2(\frac{\sqrt{6}-3\sqrt{3}}{-16}) = \frac{3\sqrt{3}-\sqrt{6}}{8}$
$(iv)\: \frac{\sin 30^{o}+\tan 45^{o}-\operatorname{cosec} 60^{o}}{\sec 30^{o}+\cos 60^{o}+\cot 45^{o}}$
Answer:
$\frac{\sin 30^{o}+\tan 45^{o}-\operatorname{cosec} 60^{o}}{\sec 30^{o}+\cos 60^{o}+\cot 45^{o}}$ ..................(i)
It is known that the values of the given trigonometric functions,
$\\\sin 30^0 = \frac{1}{2}=cos 60^0, \tan 45^0 = 1=\cot 45^0, \sec 30^0 = \frac{2}{\sqrt{3}}=\operatorname{cosec} 60^0$
Putting all these values in equation (i), we get;
$\\\Rightarrow \frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}= \frac{3\sqrt{3}-4}{3\sqrt{3}+4}\times\frac{4-3\sqrt{3}}{4-3\sqrt{3}}= \frac{12\sqrt{3}-27-16+12\sqrt{3}}{-11}= \frac{43-24\sqrt{3}}{11}$
$(v)\frac{5\cos^{2}60^{o}+ 4\sec^{2}30^{o}-\tan^{2}45^{o}}{\sin^{2}30^{o}+\cos^{2}30^{o}}$
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 :
$(iv)\frac{2\: \tan 30^{o}}{1-tan^{2}\: 30^{o}}=$
$(A)\: \cos 60^{o}$ $(B)\: \sin 60^{o}$ $(C)\: \tan 60^{o}$ $(D)\: \sin 30^{o}$
Answer:
Put the value of $tan 30^{o}={\frac{1}{\sqrt{3}}}$
$\\=\frac{2\: \tan 30^{o}}{1-tan^{2}\: 30^{o}}=\frac{2\times \frac{1}{\sqrt{3}}}{1-(\frac{1}{\sqrt{3}})^2}\\ =\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}\\ =\frac{2}{\sqrt{3}}\times \frac{3}{2}\\ =\sqrt{3}= \tan 60^0$
The correct option is (C).
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, $\\\sin(45^0+45^0) = \sin 45^0+\sin 45^0$
$\sin 90^0 = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \neq 1$
(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 $\frac{1}{\sqrt2}$ = 0.707
(iii) False,
$\cos 0^0 = 1,\ \cos 30^0 = \frac{\sqrt{3}}{2}= 0.87,\ \cos 45^0 = \frac{1}{\sqrt2}= 0.707$
(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 $\csc^2A -\cot^2A = 1$
(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,
$\\\Rightarrow \frac{1+\tan ^{2}A}{1+\frac{1}{\tan ^{2}A}}= \tan^2A\times(\frac{1+\tan^2 A}{1+\tan^2A})= \tan^2A$
$(i) (\csc \theta -\cot \theta )^{2}= \frac{1-\cos \theta }{1+\cos \theta }$
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.
$(ii) \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}= 2\sec A$
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.
$(iii)\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta$
[ 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.
$(iv)\frac{1+\sec A}{\sec A}=\frac{\sin ^{2}A}{1-\cos A}$
[ 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.
$(vi)\sqrt{\frac{1+\sin A}{1-\sin A}}= \sec A+\tan A$
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.
$(viii)(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}= 7+\tan ^{2}A+\cot ^{2}A$
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
$(ix)\:(cosec A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$
[ 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.
$(x) (\frac{1+\tan ^{2}A}{1+\cot ^{2}A})= (\frac{1-\tan A}{1-\cot A})^{2}= \tan ^{2}A$
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.
Here are the exercise-wise links for the NCERT class 10 chapter 8, Introduction to Trigonometry:
Here, students can find chapter-wise solutions.
Here are the subject-wise links for the NCERT solutions of class 10:
Given below are the subject-wise exemplar solutions of class 10 NCERT:
Here are some useful links for NCERT books and the NCERT syllabus for class 10:
This NCERT Solutions for Class 10 Maths Chapter 8 contains the basic concepts of trigonometry like trigonometric ratios, trigonometric identities, and their applications in real world.
To solve trigonometric identities in Class 10 Maths Chapter 8, use the basic trigonometric identities as sin2 θ + cos2 θ = 1 and try to express terms in sine and cosine. Also, use simplification and factorisation to solve these types of problems.
The easiest way to understand trigonometric ratios is by taking a right-angle triangle where sine, cosine, and tangent represent ratios of different sides as
sin θ = Opposite / Hypotenuse
cos θ = Adjacent / Hypotenuse
tan θ = Opposite / Adjacent
To score full marks in Class 10 Trigonometry, memorise all the important trigonometric ratios, identities, and formulas. Practice NCERT examples, exercises, and questions from the previous years. Also, regularly revise and give mock tests to improve scores.
Trigonometry is used in various fields of real life. It is mainly used for calculating the height and distance of real-life objects, and for building bridges and roads. Also, it is used in navigation systems like GPS.
Application Date:24 March,2025 - 23 April,2025
Admit Card Date:04 April,2025 - 26 April,2025
Hello
Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.
1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.
2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.
3. Counseling and Seat Allocation:
After the KCET exam, you will need to participate in online counseling.
You need to select your preferred colleges and courses.
Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.
4. Required Documents :
Domicile Certificate (proof that you are a resident of Karnataka).
Income Certificate (for minority category benefits).
Marksheets (11th and 12th from the Karnataka State Board).
KCET Admit Card and Scorecard.
This process will allow you to secure a seat based on your KCET performance and your category .
check link for more details
https://medicine.careers360.com/neet-college-predictor
Hope this helps you .
Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
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