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

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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

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]

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 \(\frac{13}{12}\).]

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°).

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.