# Maharashtra Board 11. Math Solutions Chapter 2 Trigonometry – I Ex 2.2 – Maharashtra Board Solutions (2023)

BalbhartiMaharashtra State Board Class 11 Math SolutionsPdf Chapter 2 Trigonometry – I Ex 2.2 Questions and Answers.

## Maharashtra State Board 11. Mathematical Solutions Chapter 2 Trigonometry – I Ex 2.2

Question 1.
Se A = 1 = $$\sqrt{2}$$ cos B e $$\frac{\pi}{2}$$ < A < π, $$\frac{3 \pi}{2}$$
Solution:
Given, 2sin A = 1
∴ you A = 1/2
We know,
why2A = 1 - Ohne2A = 1 – $$\left(\frac{1}{2}\right)^{2}=1-\frac{1}{4}=\frac{3}{4}$$
∴ cos A = $$\pm \frac{\sqrt{3}}{2}$$
Given that $$\frac{\pi}{2}$$ < A < π
A is in the second quadrant. We know,
sin2B = 1 - because2B = 1 – $$\left(\frac{1}{\square{2}}\right)^{2}$$$$\frac{1}{2}=\frac{1}{2}\ ) ∴ sen B = \(\pm \frac{1}{\sqrt{2}}$$
Like $$\frac{3 \pi}{2}$$ < B < 2π
B is in the fourth quadrant,  question 2
If  and A, B are angles in the second quadrant, then prove that 4cosA + 3cosB = -5
Solution:
Sim $$\frac{\sin \mathrm{A}}{3}=\frac{\sin \mathrm{B}}{4}=\frac{1}{5}$$
∴ sen A = $$\frac{3}{5}$$ y sen B = $$\frac{4}{5}$$
We know,
why2A = 1 - Ohne2= 1 – $$\left(\frac{3}{5}\right)^{2}$$ = 1 – $$\frac{9}{25}=\frac{16}{25}$$
∴Cos A = ± $$[{4}{5}$$
Since A is in the second quadrant,
because A < 0
∴ Because A = –$$\frac{4}{5}$$
Sine B = 4/5
We know,
why2B = 1 - sin2B = 1 – $$\left(\frac{4}{5}\right)^{2}=1-\frac{16}{25}=\frac{9}{25}$$
∴ Cos B = ±$$\frac{4}{5}$$
Since B is in the second quadrant, cos B < 0 question 3
If tan θ = $$\frac{1}{2}$$, then $$\frac{2\sin\theta+3\cos\theta}{4\cos\theta+3\sin\theta}$$
Solution: question 4
Remove 0 from the following:
I. x = 3secθ, y = 4tanθ
ii. x = 6 coseg θ, y = 8 coseg θ
iii. x = 4 cos θ – 5 sen θ, y = 4 sen θ + 5 cos θ
4. x = 5 + 6 cosec θ,y = 3 + 8 cot θ
v. x = 3 – 4. θ,3y = 5 + 3 Sek. θ
Solution:
I. x = 3secθ, y = 4tanθ
∴ sec θ = $$\frac{x}{3}$$ y tan θ= $$\frac{y}{4}$$
We know,
sek2θ - light brown2θ = 1 ∴ 16x2- nine years old2= 144

(Video) Trigonometry-I Ex.2.2 Part 4 Q.1-Q.4 | Class 11 Maths | Maharashtra Board | Dinesh Sir

ii. x = 6 coseg θ y y = 8 cot θ
.'. cosec θ =  y cot θ = 
We know,
cosec2θ—cuna2θ = 16x2- nine years old2= 576 iii. x = 4cos θ – 5 sen θ ... (i)
y = 4sinθ + 5cosθ .. .(ii)
If we square (i) and (ii) and add, we get
x2+j2= (4cosθ – 5sinθ)2+ (4 sen θ + 5 cos θ)2
= 16 because2θ – 40 senθ cosθ + 25 sen2θ + 16 sin2θ + 40 sen θ cos θ + 25 cos2EU
= 16 (sin2θ + cos2i) + 25(sin2θ + cos2EU)
= 16(1) + 25(1)
= 41

4. x = 5 + 6 cosec θ undy = 3 + 8cot θ
∴ x – 5 = 6 coseg θ y y – 3 = 8 coseg θ
∴ cosec θ = $$\frac{x-5}{6}$$ y cot θ = $$\frac{y-3}{8}$$
We know,
cosec2θ—cuna2θ = 1
∴ $$\left(\frac{x-5}{6}\right)^{2}-\left(\frac{y-3}{8}\right)^{2}$$ = 1

v. 2x = 3 - 4then θ and 3y = 5 + 3sec θ
∴ 2x – 3 = -4tan θ y 3y – 5 = 3seg θ
∴ tan θ = $$\frac{3-2 x}{4}$$ y seg θ = $$\frac{3 y-5}{3}$$θ
We know that, sec.2θ - light brown2θ = 1
∴ $$\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{3-2 x}{4}\right)^{2}$$ = 1
∴ $$\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{2 x-3}{4}\right)^{2}$$ = 1

question 5
say 2 sen2θ + 3sin θ = 0, find the allowed values ​​of cosθ.
Solution:
2 sin2θ + 3sen θ = 0
∴ sin θ (2 sin θ + 3) = 0
∴ sen θ = 0 o sen θ = $$\frac{-3}{2}$$
Como - 1 ≤ sin θ ≤ 1,
sen θ = 0
$$\sqrt{1-\cos^{2}\theta}$$ = 0 ...[ ∵ Som2θ = 1- pig2EU]
∴ 1 – because2θ = 0
∴ because2θ = 1
∴ cos θ = ±1 …[∵ – 1 ≤ cos θ ≤ 1]

(Video) Exercise 2.2, Q.1to 14 Trigonometry 1, class 11th maths part 1, Maharashtra board

question 6
yes 2 because2θ – 11 cos θ + 5 = 0, then find the possible values ​​of cos θ.
Solution:
2 because2θ – 11 cos θ + 5 = 0
∴ 2cos2θ – 10 cos θ – cos θ + 5 = 0
∴ 2cos θ(cos θ – 5) – 1 (cos θ – 5) = 0
∴ (cos θ – 5) (2cos θ – 1) = 0
cos θ – 5 = 0 or 2 cos θ – 1 = 0
∴ cos θ = 5 o cos θ = 1/2
Da -1 ≤ cos θ ≤ 1
∴ cos θ = 1/2

question 7
Find the acute angle θ as 2cos2θ = 3sen θ.
Solution:
2 because20 = 3 senθ
∴ 2(1 – sin2i) = 3sen i
∴2–2 sen2θ = 3sen θ
∴2sünde2θ + 3sen 9-2 = θ
∴2sünde2θ + 4sen θ – sen θ – 2 = θ
∴ 2sen θ(sen θ + 2) -1 (sen θ + 2) = θ
∴ (sen θ + 2) (2sen θ – 1) = 0
∴ sen θ + 2 = 0 o 2 sen θ – 1 = 0
∴ sen θ = -2 o sen θ = 1/2
Like, -1 ≤ sin θ ≤ 1
∴ Sen θ = 1/2
∴ θ = 30° …[ ∵ sen 30 = 1/2] question 8
Find the acute angle 0 such that 5tan2 0 + 3 = 9sec 0.
Solution:
5 tan2θ + 3 = 9 seconds θ
∴ 5(sec2θ – 1) + 3 = 9 seconds. θ
∴ 5 seconds2θ – 5 + 3 = 9 seconds. θ
∴ 5 seconds2θ – 9 seconds. θ – 2 = 0
∴ 5 seconds2i – 10 seconds. θ + sec. θ – 2 = 0
∴ 5 sec. θ(Seg. θ – 2) + 1(Seg. θ – 2) = 0
∴ (Seg. θ – 2) (5 Seg. θ + 1) = 0
∴ seg θ – 2 = 0 o 5 seg θ + 1 = 0
∴ second θ = 2 the second θ = -1/5
Given that sec θ ≥ 1 or sec θ ≤ -1,
Second. θ = 2
∴ θ = 60° … [ ∵ Seg. 60° = 2]

question 9
Encontre sen θ tal que 3cos θ + 4sin θ = 4.
Solution:
3cosθ + 4sinθ = 4
∴ 3cos θ = 4(1 – Sünde θ)
If we square both sides, we get .
9 because2θ = 16(1 – Sunde θ)2
∴ 9(1 – sin2θ) = 16(1 + sen2θ – 2sen θ)
∴9–9 domingo2θ = 16 + 16 sen2me - 32sen me
∴25sünde2θ – 32sen θ + 7 = 0
∴25sünde2θ – 25 sen θ – 7 sen θ + 7 = 0
25sen θ (sen θ – 1) – 7 (sen θ – 1) = 0
∴ (sen θ – 1) (25sen θ – 7) = 0
∴ sen θ – 1 = 0 o 25 sen θ – 7 = 0
∴ sen θ = 1 o sen θ = $$\frac{7}{25}$$
Like, -1 ≤ sin θ ≤ 1
∴ sen θ = 1 o $$\frac{7}{25}$$
[Hint: The answer given in the book is 1. However, by our calculation it is 1 or $$\frac{7}{25}$$.]

question 10
Se cosec θ + cot θ = 5, then calculate sec θ
Solution:
cosec θ + berço θ = 5
∴ $$\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=5$$
∴ $$\frac{1+\cos\theta}{\sin\theta}=5$$
∴ 1 + cosθ = 5.senθ
If we square both sides, we get
1 + 2 cos θ + cos2θ = 25sen2EU
∴ because2θ + 2 cos θ + 1 = 25 (1 – cos2EU)
∴ because2θ + 2 cos θ + 1 = 25 – 25 cos2EU
∴ 26 because2θ + 2 because θ – 24 = 0
∴ 26 because2θ + 26 cos θ – 24 cos θ – 24 = 0
∴ 26 cos θ (cos θ + 1) – 24 (cos θ + 1) = 0
∴ (cos θ + 1) (26 cos θ – 24) = 0
∴ cos θ + 1 = θ o 26 cos θ – 24 = 0
∴ cos θ = -1 o cos θ = $$\frac{24}{26}=\frac{12}{13}$$
Wenn cos θ = -1, sin θ = 0
∴ cot θ and cosec x are not defined,
∴ cosθ ≠ -1
∴ cos θ = $$\frac{12}{13}$$
∴ seg. θ = $$\frac{1}{\cos \theta}=\frac{13}{12}$$
[Note: the answer given in the book is -1 or $$\frac{13}{12}$$.
However, by our calculations, it's just $$\frac{13}{12}$$.]

(Video) Lecture 2 | Exercise 2.2 part 1 Chapter 2 Trigonometry 1 class 11 maths 1 | maharashtra board #nie

question 11
If θ = $$\frac{3}{4}$$ e π < θ < $$\frac{3 \pi}{2}$$, I find a value of 4 coseg θ + 5 cos θ.
Solution:
We know,
cosec2θ = 1 + cuna2θ = $$\left(\frac{3}{4}\right)^{2}$$ = 1 + $$\frac{9}{16}$$
∴ cosek2θ = $$\frac{25}{16}$$
∴ cosec θ = $$\pm \frac{5}{4}$$
Da π < θ < $$\frac{3 \pi}{2}$$
θ is in the third quadrant.
∴ coseg θ < 0
∴ coseg θ = –$$\frac{5}{4}$$
Cuna θ = $$\frac{3}{4}$$
tan θ = $$\frac{1}{\cot \theta}=\frac{4}{3}$$
We know,
sek2θ = 1 + tan2θ = 1 + $$\left(\frac{4}{3}\right)^{2}$$
= 1 + $$\frac{16}{9}=\frac{25}{9}$$
∴ second θ = ±$$\frac{5}{3}$$
Since θ is in the third quadrant,
Second. θ < 0
∴ second θ = –$$\frac{5}{3}$$
cos θ = $$\frac{1}{\sec \theta}=\frac{-3}{5}$$
∴ 4 coseg θ + 5 cos θ
= $$4\left(-\frac{5}{4}\right)+5\left(-\frac{3}{5}\right)$$
= -5 – 3 = -8
[Note: the question has been modified.] question 12
Find the Cartesian coordinates of the points whose polar coordinates are:
I. (3, 90°) ii. (1, 180 degrees)
Solution:
I. (r, θ) = (3, 90°)
Using x = r cos θ and y = r sin θ, where (x,y) are the necessary Cartesian coordinates, we get
x = 3cos 90° y y = 3sen 90°
∴ x = 3(0) = 0 e y = 3(1) = 3
∴ the necessary Cartesian coordinates are (0, 3).

ii. (r, θ) = (1, 180°)
Using x = r cos θ and y = r sin θ, where (x,y) are the necessary Cartesian coordinates, we get
x = 1(cos 180°) e y = 1(sen 180°)
∴ x = -1 e y = 0
∴ the necessary Cartesian coordinates are (-1, 0).

question 13
Find the polar coordinates of the points whose Cartesian coordinates are:
1. (5, 5) ii. (1, $$\square root{3}$$)
ii. (-1, -1) iv. (-$$\square root{3}$$, 1)
Solution:
I. (x, y) = (5, 5)
∴ r = $$\square root{x^{2}+y^{2}}$$ = $$\square root{25+25}$$
$$=\square root{50}=5 \square root{2}$$
tan θ = $$\frac{y}{x}=\frac{5}{5}$$ = 1
Since the given point is in the 1st quadrant,
θ = 45° …[∵ tan 45° = 1]
∴ the required polar coordinates are ($$5 \sqrt{2}$$, 45°).

ii. (x, y) = (1, $$\sqrt{3}$$)
∴ r = $$\square root{x^{2}+y^{2}}=\square root{1+3}=\square root{4}=2$$
tan θ = $$\frac{y}{x}=\frac{\sqrt{3}}{1}=\sqrt{3}$$
Since the given point is in the 1st quadrant,
θ = 60° …[∵ tan 60° = $$\sqrt{3}$$]
∴ the required polar coordinates are (2, 60°).

(Video) Class 11 Ch.2 Trigonometry-I Lecture-8 for Science/Commerce/Arts | Success Batch | Dinesh Sir

iii. (x, y) = (-1, -1)
∴ r = $$\square root{x^{2}+y^{2}}=\square root{1+1}=\square root{2}$$
tan θ = $$\frac{y}{x}=\frac{-1}{-1}=1$$
∴ tan θ = tan $$\frac{\pi}{4}$$
Since the given point is in the third quadrant,
tan θ = tan $$\left(\pi+\frac{\pi}{4}\right)$$ ...[∵ tan(n + x) = tanx]
∴ tan θ = tan $$\left(\frac{5\pi}{4}\right)$$
∴ θ = $$\frac{5 \pi}{4}$$ = 225°
∴ the required polar coordinates are ($$\sqrt{2}$$, 225°).

4. (x, y) = (-$$\sqrt{3}$$ , 1)
∴ r = $$\square root{x^{2}+y^{2}}=\square root{3+1}=\square root{4}=2$$
tan θ = $$\frac{y}{x}=\frac{1}{-\sqrt{3}}$$ = -tan $$\frac{\pi}{6}$$
Since the specified point is in the second quadrant,
tan θ = tan $$\left(\pi-\frac{\pi}{6}\right)$$ ...[∵ tan (π – x) = – tanx]
∴ tan θ = tan $$\left(\frac{5\pi}{6}\right)$$
∴ θ = $$\frac{5 \pi}{6}$$ = 150°
∴ the required polar coordinates are (2, 150°)

question 14
Find the values ​​of:
I. Sin$$\frac{19 \pi^{e}}{3}$$
ii. Korb 1140°
iii. Cuna $$\frac{25 \pi^{e}}{3}$$
Solution:
I. We know that the sine function is periodic with period 2π.
sin $$\frac{19 \pi}{3}$$ = sin $$\left(6 \pi+\frac{\pi}{3}\right)$$ = sin $$\frac{\pi}{ 3}=\frac{\sqrt{3}}{2}$$

ii. We know that the cosine function is periodic with period 2π.
cos 1140° = cos (3 × 360° + 60°)
= cos 60° = $$\frac{1}{2}$$

iii. We know that the cotangent function is periodic with period π.
Children's bed $$\frac{25 \pi}{3}$$ = Children's bed $$\left(8 \pi+\frac{\pi}{3}\right)$$ = Children's bed $$\frac{\ pi}{ 3}$$ = $$\frac{1}{\sqrt{3}}$$
dhana work.txt
See the work of dhana.txt.

## FAQs

### Is Class 11 trigonometry hard? ›

In Class 11 Trigonometric Functions is the easiest one among all other chapters. Though there are a number of formulas, students can easily score the marks for this chapter.

Which chapter is trigonometry in class 11? ›

NCERT Solutions Class 11 Maths Chapter 3 Trigonometric Functions- 2022-23 Free PDF Download.

Is there trigonometry in class 11 Maths? ›

So basically, trigonometry is a study of triangles, which has angles and lengths on its side. Trigonometry basics consist of sine, cosine and tangent functions. Trigonometry for class 11 contains trigonometric functions, identities to solve complex problems more simply.

How many chapters are there in Maths class 11 Maharashtra board? ›

Answer: The current 11th Maths and Statistics syllabus Maharashtra board includes 18 chapters which are divided into two parts.

Is Class 11 hard or 12? ›

Hi 11th standard is comparatively tougher than 12th. so before starting 11th class go through the basics and formulas related to 11th class and try to solve questions of math, physics, chemistry from NCERT book.

Is trigonometry easy or hard? ›

Trigonometry is considered hard by students due to the following reasons: The memorization of formulas and values. Non-linear functions. Angle measurement in Radians/Degree.

What is the hardest chapter in class 11 Maths? ›

Circle, Parabola and Permutation and Combination are tough chapters of Class 11 Maths. Sequence and Series is another tough topic that needs more attention and preparation. Additionally, you must also study Coordinate Geometry and Integral Calculus for JEE Mains 2022.

Which chapters are important in Maths 11? ›

Important Class 11 Maths Topics from Each Chapter
• Cartesian Product of Sets.
• Concept of Relations.
• Real-Valued Functions.

Is trigonometry an important chapter? ›

Trigonometry is important to mathematics as it involves the study of calculus, statistics and linear algebra. This article covers the identities of trigonometry and trigonometric equations, functions of trigonometry, properties of inverse trigonometric functions, problems on heights and distances.

Can we skip Maths in Class 11? ›

To pursue a career in these fields, you are not required to be a genius in Mathematics. However, a good hold over numbers is surely required. Knowledge of statistics and analytical skills are a must, and thus maths till class 12th can be beneficial, even if it is not compulsory.

### How do you solve trigonometry questions easily? ›

Make a point of memorizing them.
1. Quotient Identities: tan(x) = sin(x)/cos(x) cot(x) = cos(x)/sin(x)
2. Reciprocal Identities: csc(x) = 1/sin(x) sec(x) = 1/cos(x) cot(x) = 1/tan(x) sin(x) = 1/csc(x) ...
3. Pythagorean Identities: sin2(x) + cos2(x) = 1. cot2A +1 = csc2A. 1+tan2A = sec2A.

What grade math is trigonometry? ›

In general, trigonometry is taken as part of sophomore or junior year math. In addition to being offered as its own course, trigonometry is often incorporated as a unit or semester focus in other math courses.

Which chapter has more weightage in Maths class 11? ›

Algebra is the highest weightage unit of the CBSE 11th Maths syllabus which is followed by Sets and Function with the weightage of 23 marks.

Which chapters are removed from class 11? ›

CBSE Class 11 History Deleted Syllabus 2023: FAQs

The chapters which have been removed from Class 11 history syllabus are Early Societies, Nomadic Empires, and Confrontation of Cultures. Will syllabus be reduced for Class 11? Yes syllabus has been reduced form Class 11 history for the academic year 2021-2022.

Which topics are removed from Maths 11? ›

Maths Class 11 Deleted Syllabus 2022-23

Difference of sets. General Solutions of trigonometric equations of the type sin y = sin a, cos y = cos a, and tan y = tan a. Polar representation of complex numbers. Square root of a complex number.

Why is 11th so hard? ›

11th is very different in terms of the course content as compared to what you have studied till 10th. Difficulty level is higher if you have chosen the science stream as now it is no more general science. 11th grade will teach you concepts of fundamental physics, chemistry, biology, maths etc.

Can you fail class 11? ›

Totally depends on your school/college. They might take a re-test or just promote you directly to 12th. There is nothing to worry about. They never let you fail you in class 11th.

How to pass in class 11? ›

Although the basic criteria to pass class 11 in each subjects of the science stream is about to score at least 21 in all of the subjects out of 70, as per the state board marking rules. However in other boards as well in class 11 you need to score at least 30% marks in each and every subjects.

How can I pass trigonometry? ›

How to Pass Trigonometry
1. Review What You Have Learned in Previous Classes. Before beginning your trig class, make sure you start off on solid footing. ...
2. Be a Star Student. Strive to attend every class. ...
3. Trig Doesn't End at the Classroom Door. ...
5. Making Time. ...

What's the hardest math? ›

The Riemann Hypothesis, famously called the holy grail of mathematics, is considered to be one of the toughest problems in all of mathematics.

### How do you study trigonometry fast? ›

What is the Easiest Way to Learn Trigonometry?
1. Study all the basics of trigonometric angles.
2. Study right-angle triangle concepts.
3. Pythagoras theorem.
4. Sine rule and Cosine rule.
5. List all the important identities of trigonometry.
6. Remember the trigonometry table.
7. Be thorough with the trigonometric formulas.

Which is the toughest chapter in maths class 11? ›

Circle, Parabola and Permutation and Combination are tough chapters of Class 11 Maths. Sequence and Series is another tough topic that needs more attention and preparation. Additionally, you must also study Coordinate Geometry and Integral Calculus for JEE Mains 2022.

Which is difficult subject in 11th? ›

Science (PCM or PCB)

Science stream has a reputation of most difficult stream in class 11th and 12th. Students learn Science till 10th standard, so they are aware of the basics of this subject.

Is trigonometry a hard class? ›

Trigonometry is hard because it deliberately makes difficult what is at heart easy. We know trig is about right triangles, and right triangles are about the Pythagorean Theorem. About the simplest math we can write is When this is the Pythagorean Theorem, we're referring to a right isosceles triangle.

Is trigonometry harder than precalculus? ›

Consequently, some students have a steep learning curve upon entering precal and will feel like they are swimming in unchartered waters for a while. Now, most students agree that math analysis is “easier” than trigonometry, simply because it's familiar (i.e., it's very similar to algebra).

Why is 11th standard hard? ›

11th is very different in terms of the course content as compared to what you have studied till 10th. Difficulty level is higher if you have chosen the science stream as now it is no more general science. 11th grade will teach you concepts of fundamental physics, chemistry, biology, maths etc.

Which is the easiest chapter in maths class 11? ›

Generally easiness of chapter in Mathematics depends upon the priority of a person's interest. However some chapters like set theory, mathematical induction, A.P, G.P, differentiation, matrix in class 11 and 12 are considered to less complicated compared to other chapters.

Which chapter is most important in maths class 11? ›

Important Class 11 Maths Topics from Each Chapter
• Cartesian Product of Sets.
• Concept of Relations.
• Real-Valued Functions.

What is the most failed course in high school? ›

Algebra is the single most failed course in high school, the most failed course in community college, and, along with English language for nonnative speakers, the single biggest academic reason that community colleges have a high dropout rate.

Which is the most difficult grade in school? ›

While junior year is often the hardest year of high school, the transition from middle school to 9th grade can also be tough. To make it easier, don't feel afraid to reach out to your teachers and counselors, and take advantage of the support resources that are available.

### Which subject is hardest for students? ›

The hardest degree subjects are Aerospace Engineering, Law, Chartered Accountancy, Architecture, Chemistry, Medicine, Pharmacy, Psychology, Statistics, Nursing, Physics, Astrophysics, Biomedical Engineering, Astronomy, and Dentistry. Let's dive right in, and look at why these courses are the hardest degree subjects.

Which math class is hardest? ›

What is the Hardest Math Class in High School? In most cases, you'll find that AP Calculus BC or IB Math HL is the most difficult math course your school offers. Note that AP Calculus BC covers the material in AP Calculus AB but also continues the curriculum, addressing more challenging and advanced concepts.

What grade level is trigonometry? ›

In general, trigonometry is taken as part of sophomore or junior year math. In addition to being offered as its own course, trigonometry is often incorporated as a unit or semester focus in other math courses.

Is calculus or trigonometry easier? ›

BTW basic calculus is not easier than basic trigonometry because basic trigonometric essentially deals with algebraic properties of circular functions. Calculus on the other hand is essentially non-algebraic in nature.

Is trigonometry the most important math? ›

As one of the most important branches of mathematics, trigonometry is something that every student should focus on. Great trigonometry skills allow students to work out complex angles and dimensions in relatively little time.

## Videos

1. Trigonometry I Ex.2.2 Part 6 | Class 11 Maths | Maharashtra Board | Dinesh Sir
(DINESH SIR Live Study)
2. Proof of identities in hindi | Trigonometry 1 exercise 2.2 class 11 maths question 15 hsc board
(SHALINI J KUSHWAHA { ALL STUDY })
3. Trigonometry 1 Exercise 2.2 Class 11 Maharashtra Board New Syllabus Part - 8
4. 11th std Trigonometry 1 maths 1 Exercise 2.2 Part 1 Class 11 Trigonometry 1 Exercise 2.2 Q.1 to Q.6
(Suhani S. Suryawanshi)
5. Lecture 3 | Exercise 2.2 part 2 Chapter 2 Trigonometry 1 class 11 maths 1 | maharashtra board #nie
(New Indian Era (NIE) - Prashant Tiwari)
6. Exercise 2.2 class 11 maths trigonometry part 1 lecture no 6 maharashtra board
(SHALINI J KUSHWAHA { ALL STUDY })
Top Articles
Latest Posts
Article information

Author: Gregorio Kreiger

Last Updated: 01/30/2023

Views: 6839

Rating: 4.7 / 5 (77 voted)

Author information

Name: Gregorio Kreiger

Birthday: 1994-12-18

Address: 89212 Tracey Ramp, Sunside, MT 08453-0951

Phone: +9014805370218

Job: Customer Designer

Hobby: Mountain biking, Orienteering, Hiking, Sewing, Backpacking, Mushroom hunting, Backpacking

Introduction: My name is Gregorio Kreiger, I am a tender, brainy, enthusiastic, combative, agreeable, gentle, gentle person who loves writing and wants to share my knowledge and understanding with you.