Important question-chapter 3,4

Download Tamil Nadu 11th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

Questions:

Generic 5

6 x 5 = 30

Suppose two radar stations located 100 km apart, each detect a fighter aircraft between them. The angle of elevation measured by the first station is 30°, whereas the angle of elevation measured by the second station is 45°. Find the altitude of the aircraft at that instant.

In \(\triangle\)ABC, Prove the following
\(\frac { asin(B-C) }{ { b }^{ 2 }-{ c }^{ 2 } } =\frac { bsin(C-A) }{ { c }^{ 2 }-{ a }^{ 2 } } =\frac { csin(A-B) }{ { a }^{ 2 }-{ b }^{ 2 } } \)

Prove that cos (A + B) cos C - cos (B + c) cos A = sin B sin (C - A)

If tan² \(\theta\) = 1 - k², Show that sec \(\theta\) + tan³ \(\theta\) cosec \(\theta\) = (2 -k²)^3/2. Also, find the value of k for which this result holds

Two vehicles leave the same place P at the same time moving along two different roads. One vehicle moves at an average speed of 60 km/hr and the other vehicle moves at an average speed of 80 km/hr. After half an hour the vehicle reach the destinations A and B. If AB subtends 60^o at the initial point P, then find AB.

A fighter jet has to hit a small target by flying a horizontal distance. When the target is sighted, the pilot measures the angle of depression to be 30⁰. If after 100 km, the target has an angle of depression of 60⁰, how far is the target from the fighter jet at that instant?

Generic 3

6 x 3 = 18

Find the value of \(sin\left( 22\frac { 1^{ 0 } }{ 2 } \right) \)

Prove that \(\tan { \left( \frac { \pi }{ 4 } +\theta \right) } -\tan { \left( \frac { \pi }{ 4 } -\theta \right) } =2\tan { 2\theta } \)

If tan x = \(\frac{n}{n+1}\) and tan y = \(\frac{1}{2n+1}\), find tan (x + y).

10.

Prove that \(\frac { sin(4A-2B)+sin(4B-2A) }{ cos(4A-2B)+cos(4B-2A) } =tan(A+B)\)

11.

Prove that \(\frac { sin4x+sin2x }{ cos4x+cos2x } =tan3x\)

12.

If sin A = \(\frac{3}{5}\) and cos B = \(\frac{9}{41}\), 0 < A < \(\frac{\pi}{2}\), 0 < B < \(\frac{\pi}{2}\). Find the value of cos (A - B)

Generic 2

6 x 2 = 12

13.

Express each of the following product as a sum or difference. cos 110° sin 55°.

14.

Find the values of sin (-45°).

15.

Express each of the following as a product.
cos 65^o + cos 15^o

16.

Find the values of cos 2A, A lies in the first quadrant, when cos A = \(\frac{15}{17}\)

17.

Find the degree measure corresponding to the following radian measure; \(\frac { 2\pi }{ 5 } \)

18.

Express each of the following angles in radian measure
135⁰

Multiple Choice Question

10 x 1 = 10

19.

Area of triangle ABC is _______________

(a)

\(\frac{1}{2}\)ab cos C

(b)

\(\frac{1}{2}\)ab sin C

(c)

\(\frac{1}{2}\)ab cos B

(d)

\(\frac{1}{2}\)bc sin B

20.

If 2 sin θ + 1 = 0 and \(\sqrt{3}\) tan θ = 1, then the most general value of θ is _______________

(a)

\(n\pi\pm\frac{\pi}{6}\)

(b)

\(n\pi+(-1)^n\frac{7\pi}{6}\)

(c)

\(2n\pi+\frac{7\pi}{6}\)

(d)

\(2n\pi+\frac{11\pi}{6}\)

21.

In \(\triangle\)ABC, \(\hat{C}\) = 90° then a cos A + b cos B is _______________

(a)

2R sin B

(b)

2 sin B

(c)

(d)

2a sin B

22.

sin²\((22{1\over 2}^o)\) is ____________

(a)

\({\sqrt{2-\sqrt{2}}}\over2\)

(b)

\({2\sqrt{2}-1\over 4\sqrt{2}}\)

(c)

\({\sqrt{2-\sqrt{2}\over 2}}\)

(d)

none of these

23.

The product of first n odd natural numbers equals

(a)

²ⁿC_n\(\times\)ⁿP_n

(b)

\({ \left( \frac { 1 }{ 2 } \right) }^{ n }\times\)²ⁿC_n\(\times\)ⁿP_n

(c)

\({ \left( \frac { 1 }{ 4 } \right) }^{ n }\times \)²ⁿC_n\(\times\)²ⁿP_n

(d)

ⁿC_n\(\times\)ⁿP_n

24.

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines.

(a)

(b)

(c)

(d)

25.

cos 35⁰+ cos 85⁰+ cos 155⁰= _______________

(a)

(b)

\(\frac { 1 }{ \sqrt { 3 } } \)

(c)

\(\frac { 1 }{ \sqrt { 2 } } \)

(d)

cos 275⁰

26.

\(\frac { cos6x+6cos4x+15cos2x+10 }{ cos5x+5cos3x+10cosx } \) is equal to

(a)

cos 2x

(b)

cos x

(c)

cos 3x

(d)

2 cos x

27.

If cos 28⁰+ sin 28⁰= k³, then cos 17⁰ is equal to

(a)

\(\frac { { k }^{ 3 } }{ \sqrt { 2 } } \)

(b)

-\(\frac { { k }^{ 3 } }{ \sqrt { 2 } } \)

(c)

±\(\frac { { k }^{ 3 } }{ \sqrt { 2 } } \)

(d)

-\(\frac { { k }^{ 3 } }{ \sqrt { 3 } } \)

28.

\(\frac { 1 }{ cos{ 80 }^{ 0 } } -\frac { \sqrt { 3 } }{ sin{ 80 }^{ 0 } } \)=

(a)

\(\sqrt{2}\)

(b)

\(\sqrt{3}\)

(c)

(d)

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 5

6 x 5 = 30

Let R₁ and R₂ be two radar stations and A be the position of fighter aircraft at the time of detection.
Let x be the required altitude of the aircraft.

Draw \(\bot AN\) from A to R₁R₂ meeting at N.
\(\angle A=180^0-(30^0+45^0)=105^0\)
Thus, \(\frac{a}{sin\ 45^0}=\frac{100}{sin\ 105^0}\) \(\Rightarrow a=\frac{100}{\frac{\sqrt 3+1}{2\sqrt 2}}\times\frac{1}{\sqrt 2}=\frac{200(\sqrt 3-1)}{2}\)\(=100\times(\sqrt 3-1)km\)
Now, \(sin\ 30^0=\frac{x}{a}\Rightarrow x=50\times(\sqrt 3-1)km\)

Let \(\frac { a }{ sinA } =\frac { b }{ sin B } =\frac { c }{ sinC } =k\)
a = k sin A, b = k sin B, c = k sin C...(1)
Consider \(\frac { asin\left( B-C \right) }{ { b }^{ 2 }-{ c }^{ 2 } } =\frac { ksinAsin\left( B-C \right) }{ { k }^{ 2 }{ sin }^{ 2 }B-{ k }^{ 2 }{ sin }^{ 2 }C } \)
\(\frac { ksin\left( B+C \right) sin\left( B-C \right) }{ { k }^{ 2 }\left( { sin }^{ 2 }B-{ sin }^{ 2 }C \right) } =\frac { k\left( { sin }^{ 2 }B-{ sin }^{ 2 }C \right) }{ { k }^{ 2 }\left( { sin }^{ 2 }B-{ sin }^{ 2 }C \right) } =\frac { 1 }{ k } \)...(2)
\(\frac { bsin\left( C-A \right) }{ { c }^{ 2 }-{ a }^{ 2 } } =\frac { ksin B\quad sin\left( C-A \right) }{ { k }^{ 2 }{ sin }^{ 2 }C-{ k }^{ 2 }{ sin }^{ 2 }A } =\frac { sin\left( C+A \right) \quad sin\left( C-A \right) }{ ksin\left( C+A \right) \quad sin\left( C-A \right) } =\frac { 1 }{ k } \)...(3)
Similarly, \(\frac { Csin\left( A-B \right) }{ { a }^{ 2 }-{ b }^{ 2 } } =\frac { 1 }{ k } \)...(4)
From (2), (3) and (4)
\(\frac { asin\left( B-C \right) }{ { b }^{ 2 }-{ c }^{ 2 } } =\frac { bsin\left( C-A \right) }{ { c }^{ 2 }-{ a }^{ 2 } } =\frac { csin\left( A-B \right) }{ { a }^{ 2 }-{ b }^{ 2 } } \)

LHS = cos (A + B) cos C - cos (B + C) cos A
= (cos A cos B - sin A sin B) cos C - cos A (cos B cos C - sin B sin C)
= cos A cos B cos C - sin A sin B cos C - cos A cos B cos C + cos A sin B sin C
= cos A sin B sin C - sin A sin B cos C
= sin B (cos A sin C - sin A cos C)
= sin B (sin C cos A - cos C sin A)
= sin B sin (C - A)
= RHS
Hence proved.

Given tan²θ = 1- k²
Adding 1 both sides we get,
1 + tan²θ = 1 + 1-k²
⇒ sec²θ = 2-k²
Taking power 3/2 both sides we get,
(sec²θ)^3/2= (2-k²)^3/2
⇒ sec³θ = (2-k²)^3/2
⇒ sec θ sec²θ = (2-k²)^3/2
⇒ sec θ + sec θ tan²θ = (2-k²)^3/2
⇒ sec θ +\(\frac{1}{cos\theta}.\frac{sin^2\theta}{cos^2\theta}\) = (2-k²)^3/2
⇒ sec θ + tan θ.cosec θ.tan²θ = (2-k²)^3/2
⇒ sec \(\theta\) + cosec \(\theta\) tan³\(\theta\) = (2- k²)^3/2
Hence proved.

Speed taken by the vehicle 1 = 60 km/hr.
Time \(={1\over 2}hr\)
∴ Distance = speed x time \(=\left(60\times{1\over 2}\right)=30km⇒PA=30km\)
Speed take by the 11^th vehicle = 80km/hr
Time \(={1\over 2}hr\)
∴ Distance = speed x time = 80 x \(1\over 2\) = 40km ⇒ PB = 40
Given ㄥAPB = 60⁰

Using cosine formula, c² - = a² + b² - 2ab cos C.
⇒ c² = 40² + 30² - 2(40)(30) cos 60°
c² = 160+900-2(40)(30)(1/2)
= 2500 - (40) (30) = 1300
\(⇒\ c=\sqrt{1300}=\sqrt{13\times100}=10\sqrt{13}km\)

Let C be the position of the target and A and B be the positions of the fighter jet

Given ㄥBAC = 30, ㄥABC = 45
ஃ ㄥC = 180 - (30 - 45) = 180 - 75 = 105
Given AB = 100 km
Using sine formula,
\(\frac { a }{ sinA } =\frac { c }{ sinC } \)
\(\Rightarrow \frac { a }{ sin30° } =\frac { 100 }{ sin105° } \)
\(\Rightarrow \frac { a }{ \frac { 1 }{ 2 } } =\frac { 100 }{ sin105° } \Rightarrow 2a=\frac { 100 }{ sin105° } \Rightarrow a=\frac { 50 }{ sin105° } \)
Now, sin 105° = sin (60 + 45) = sin 60 cos 45 + cos 60 sin 45
= \(\frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } .\frac { 1 }{ \sqrt { 2 } } =\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \)
Substituting (2) in (1) we get,
a = \(\frac { 50 }{ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } } \Rightarrow a=\frac { 50\left( 2\sqrt { 2 } \right) }{ \sqrt { 3 } +1 } =\frac { 100\sqrt { 2 } }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \)
a = \(\frac { 100\left( \sqrt { 6 } -\sqrt { 2 } \right) }{ 3-1 } =50\left( \sqrt { 6 } -\sqrt { 2 } \right) km\)

Generic 3

6 x 3 = 18

We know that \(cos\theta =1-2{ sin }^{ 2 }\frac { \theta }{ 2 } \Rightarrow { sin }\frac { \theta }{ 2 } =\pm \sqrt { \frac { 1-cos\theta }{ 2 } } \), Take θ = 45°
we get \(sin\frac { { 45 }^{ 0 } }{ 2 } =\pm \sqrt { \frac { 1-cos{ 45 }^{ 0 } }{ 2 } } \)(taking positive sign only, since \(22\frac { 1^{ 0 } }{ 2 } \) lies in the first quadrant)
Thus, \(sin22\frac { 1^{ 0 } }{ 2 } =\sqrt { \frac { 1-\frac { 1 }{ \sqrt { 2 } } }{ 2 } } =\frac { \sqrt { 2-\sqrt { 2 } } }{ 2 } \)

LHS = \(\tan { \left( \frac { \pi }{ 4 } +\theta \right) } -\tan { \left( \frac { \pi }{ 4 } -\theta \right) } \)
\(=\frac { \tan { \left( \frac { \pi }{ 4 } -\theta \right) +\tan { \theta } } }{ 1-\tan { \frac { \pi }{ 4 } } .\tan { \theta } } -\frac { \tan { \frac { \pi }{ 4 } } -\tan { \theta } }{ 1+\tan { \frac { \pi }{ 4 } } .\tan { \theta } } \)
= \(\frac { 1+\tan { \theta } }{ 1-\tan { \theta } } -\frac { 1-\tan { \theta } }{ 1+\tan { \theta } } \)
= \(\frac { 4\tan { \theta } }{ 1-\tan ^{ 2 }{ \theta } } =\frac { 2\times 2\tan { \theta } }{ 1-\tan ^{ 2 }{ \theta } } =2\tan { 2\theta } \)
= RHS
Hence proved.

Given tan x = \(\frac{n}{n+1}\) and tan y = \(\frac{1}{2n+1}\)
\(\tan { \left( x+y \right) } =\frac { \tan { x } +\tan { y } }{ 1-\tan { x } ,\tan { y } } \)
\(=\frac { { 2n }^{ 2 }+2n+1 }{ { 2n }^{ 2 }+n+2n+1-n } =\frac { { 2n }^{ 2 }+2n+1 }{ { 2n }^{ 2 }+2n+1 } =1\)
\(\therefore\tan { \left( x+y \right) } =1\)

10.

\(LHS=\frac { sin(4A-2B)+sin(4B-2A) }{ cos(4A-2B)+cos(4B-2A) } \)
\(=\frac { 2sin\left( \frac { 4A-2B+4B-2A }{ 2\qquad } \right) cos\left( \frac { 4A-2B-4B+2A }{ 2 } \right) }{ 2cos\left( \frac { 4A-2B+4B-2A }{ 2\qquad } \right) cos\left( \frac { 4A-2B-4B+2A }{ 2 } \right) } =\frac { sin\left( \frac { 2A+2B }{ 2 } \right) cos\left( \frac { 6A-6B }{ 2 } \right) }{ cos\left( \frac { 2A+2B }{ 2 } \right) cos\left( \frac { 6A-6B }{ 2 } \right) } \)
\(=\frac { sin(A+B) }{ cos(A+B) } =tan(A+B)=RHS.\)

11.

\(LHS=\frac { sin4x+sin2x }{ cos4x+cos2x } \)
\(=\frac { 2sin\left( \frac { 4x+2x }{ 2 } \right) .cos\left( \frac { 4x-2x }{ 2 } \right) }{ 2cos\left( \frac { 4x+2x }{ 2 } \right) .cos\left( \frac { 4x-2x }{ 2 } \right) } =\frac { sin3x.cosx }{ cos3x.cosx } =tan3x\) = RHS

12.

Since 0\(\frac { \pi }{ 2 } \) and 0\(\frac { \pi }{ 2 } \), A, B lies in the I quadrant
ஃ All the trigonometric ratios are positive

= \(\sqrt { { 5 }^{ 2 }-{ 3 }^{ 2 } } =4\)

= \(\sqrt { { 41 }^{ 2 }-{ 9 }^{ 2 } } \)
= 40
\(sinA=\frac { 3 }{ 5 } \quad sinB=\frac { 40 }{ 41 } \)
\(cosA=\frac { 4 }{ 5 } \quad cosB=\frac { 9 }{ 41 } \)
cos (A - B) = cosA cosB + sinAsinB
= \(\frac { 4 }{ 5 } .\frac { 9 }{ 41 } +\frac { 3 }{ 5 } .\frac { 40 }{ 41 } \)
= \(\frac { 36 }{ 205 } +\frac { 120 }{ 205 } =\frac { 156 }{ 205 } \)

Generic 2

6 x 2 = 12

13.

We know that 2 cos A sin B = sin (A + B) - sin (A - B)
Take A = 110° and B = 55°. We get,
2 cos 110° sin 55° = sin (110° + 55°) - sin (110° - 55°).
Thus, \(cos110^0sin55^0=\frac{1}{2}[sin165^0-sin55^0]\)

14.

sin (-45°)
= - sin(45°) =\(-\frac{1}{\sqrt 2}.\)

15.

cos 65^o + cos 15^o = \(2\cos { \left( \frac { 65+15 }{ 2 } \right) } .\cos { \left( \frac { 65-15 }{ 2 } \right) } \)
= 2 cos 40^o cos 25^o

16.

cos A = \(\frac{15}{17}\)
cos 2A = 2 cos² A - 1
= 2 \({ \left( \frac { 15 }{ 17 } \right) }^{ 2 }-1\)
= 2 \( (\frac { 225 }{ 289 })-1\)
= \( \frac { 450 }{ 289 }-1=\frac{450-289}{289}=\frac{161}{289}\)

17.

\(\frac { 2\pi }{ 5 } \)
\(\frac { 2\pi }{ 5 } \) \(\times\) \(\frac { 180 }{ \pi } \) = 2 \(\times\) 36⁰ = 72⁰

18.

135⁰
135⁰ = 135\(\times\) \(\frac { \pi }{ 180 } =\frac { 3\pi }{ 4 } \)

Multiple Choice Question

10 x 1 = 10

19.

(b)
\(\frac{1}{2}\)ab sin C

20.

21.

(d)
2a sin B

22.

(b)
\({2\sqrt{2}-1\over 4\sqrt{2}}\)

23.

\(1.3 .5 \ldots .(2 n-1) =\frac{1.2 .3 .4 \ldots \ldots .(2 n-1)(2 n)}{2.4 \ldots . \ldots(2 n)} \)
\(=\frac{\lfloor 2 n}{2^{n}\lfloor n}=\left(\frac{1}{2}\right)^{n} \cdot{ }^{2 n} C_{n} \times{ }^{n} P_{n} \)

24.

Number of parallelograms
\(={ }^{4} C_{2} \times{ }^{3} C_{2}=6 \times 3=18\)

25.

(a)
0

26.

\(\text { Numerator }=\cos 6 x+6 \cos 4 x+15 \cos 2 x+10\)
\(=(\cos 6 x+\cos 4 x)+5(\cos 4 x+\cos 2 x)+ 10(\cos 2 x+1) \)
\(=2 \cos 5 x \cos x+5(2 \cos 3 x \cdot \cos x)+10\left(2 \cos ^{2} x\right) \)
\(=2 \cos x[\cos 5 x+5 \cos 3 x+10 \cos x] \)
\(\therefore \frac{\mathrm{Nr}}{\mathrm{Dr}} =\frac{2 \cos x(\cos 5 x+5 \cos 3 x+10 \cos x)}{\cos 5 x+5 \cos 3 x+10 \cos x} \)
\(=2 \cos x \)

27.

\(\cos 28^{\circ}+\sin 28^{\circ} =\mathrm{k}^{3} \)
\(\cos 28^{\circ}+\sin \left(90^{\circ}-62^{\circ}\right) =\mathrm{k}^{3} \)
\(\cos 28^{\circ}+\cos 62^{\circ} =\mathrm{k}^{3} \)
\(2 \cos 45^{\circ} \cos 17^{\circ} =\mathrm{k}^{3} \)
\(2 \frac{1}{\sqrt{2}} \cos 17^{\circ} =\mathrm{k}^{3} \)
\(\cos 17^{\circ} =\frac{\mathrm{k}^{3}}{\sqrt{2}} \)

28.

\(x =\frac{1}{\cos 80^{\circ}}-\frac{\sqrt{3}}{\sin 80^{\circ}} \)
\(=\frac{\sin 80^{\circ}-\sqrt{3} \cos 80^{\circ}}{\sin 80^{\circ} \cos 80^{\circ}} \)
\(\frac{x}{2} =\frac{\frac{1}{2} \sin 80^{\circ}-\frac{\sqrt{3}}{2} \cos 80^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 80^{\circ} \cos 60^{\circ}-\cos 80^{\circ} \sin 60^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 20^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{\sin 160^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}} \)
\(=\frac{2 \sin 80^{\circ} \cos 80^{\circ}}{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}}=4 \)