# Trigonometry Handwritten Notes PDF By Ankur Yadav

## Trigonometry Handwritten Notes PDF By Ankur Yadav

Hello Students,

Trigonometry Notes in Hindi is most important for all Sarkari exam. so today we Provide you Complete Maths Notes PDF with important tricks. This Maths Book in Hindi and English PDF for SSC, UPSC, Railway and All Competitive Exams. We Know That all Sarkari Exams are started after a few months. In those exams, / Many Questions are coming fromMaths. so  Ankur Yadav Math Handwriting Notes in Hindi for SSC, UPSC, Railway and All Competitive Exams is important in all GOVT. Exams. Common questions are placed in Ankur Yadav Handwritten Notes PDF for SSC, UPSC, Railway and All Competitive Exams, which has been put together in most examinations, you can download, which has been put together in most examinations, you can download this trigonometry handwriting notes pdf by ankur yadav pdf very simply by clicking on the Download Button at the bottom.

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Every exam you will get at least 20-25 questions from Maths. So candidates must focus on trigonometry notes pdf and download this Maths Book to get important questions with the best solutions. We have put all Previous Year Questions of Maths that are Asked in various Govt & Private Exam.

1. From the given figure, find tan P – cot R.

Solution:

From the given,

PQ = 12 cm

PR = 13 cm

In the right triangle PQR, Q is right angle.

By Pythagoras theorem,

PR2 = PQ2 + QR2

QR2 = (13)2 – (12)2

= 169 – 144

= 25

QR = 5 cm

tan P = QR/PQ = 5/12

cot R = QR/PQ = 5/12

So, tan P – cot R = (5/12) – (5/12) = 0

 Trigonometric ratios of complementary angles: sin (90° – A) = cos A cos (90° – A) = sin A tan (90° – A) = cot A cot (90° – A) = tan A sec (90° – A) = cosec A cosec (90° – A) = sec A Trigonometric identities: cos2A + sin2A = 1 1 + tan2A = sec2A cot2A + 1 = cosec2A

2. Prove that (sin4θ – cos4θ +1) cosec2θ = 2

Solution:

L.H.S. = (sin4θ – cos4θ +1) cosec2θ

= [(sin2θ – cos2θ) (sin2θ + cos2θ) + 1] cosec2θ

Using the identity sin2A + cos2A = 1,

= (sin2θ – cos2θ + 1) cosec2θ

= [sin2θ – (1 – sin2θ) + 1] cosec2θ

= 2 sin2θ cosec2θ

= 2 sin2θ (1/sin2θ)

= 2

= RHS

3. Prove that (√3 + 1) (3 – cot 30°) = tan360° – 2 sin 60°.

Solution:

LHS = (√3 + 1)(3 – cot 30°)

= (√3 + 1)(3 – √3)

= 3√3 – √3.√3 + 3 – √3

= 2√3 – 3 + 3

= 2√3

RHS = tan360° – 2 sin 60°

= (√3)3 – 2(√3/2)

= 3√3 – √3

= 2√3

Therefore, (√3 + 1) (3 – cot 30°) = tan360° – 2 sin 60°.

Hence proved.

4. If tan(A + B) = √3 and tan(A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B.

Solution:

tan(A + B) = √3

tan(A + B) = tan 60°

A + B = 60°….(i)

And

tan(A – B) = 1/√3

tan(A – B) = tan 30°

A – B = 30°….(ii)

A + B + A – B = 60° + 30°

2A = 90°

A = 45°

Substituting A = 45° in (i),

45° + B = 60°

B = 60° – 45° = 15°

Therefore, A = 45° and B = 15°.

5. If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

Solution:

Given,

sin 3A = cos(A – 26°); 3A is an acute angle

cos(90° – 3A) = cos(A – 26°) {since cos(90° – A) = sin A}

⇒ 90° – 3A = A – 26

⇒ 3A + A = 90° + 26°

⇒ 4A = 116°

⇒ A = 116°/4

⇒ A = 29°

6. If A, B and C are interior angles of a triangle ABC, show that sin (B + C/2) = cos A/2.

Solution:

We know that, for a given triangle, the sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

B + C = 180° – A

Dividing both sides of this equation by 2, we get;

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Take sin on both sides,

sin (B + C)/2 = sin (90° – A/2)

⇒ sin (B + C)/2 = cos A/2 {since sin(90° – x) = cos x}

7. If tan θ + sec θ = l, prove that sec θ = (l2 + 1)/2l.

Solution:

Given,

tan θ + sec θ = l….(i)

We know that,

sec2θ – tan2θ = 1

(sec θ – tan θ)(sec θ + tan θ) = 1

(sec θ – tan θ) l = 1 {from (i)}

sec θ – tan θ = 1/l….(ii)

tan θ + sec θ + sec θ – tan θ = l + (1/l)

2 sec θ = (l2 + 1)l

sec θ = (l2 + 1)/2l

Hence proved.

8. Prove that (cos A – sin A + 1)/ (cos A + sin A – 1) = cosec A + cot A, using the identity cosec2A = 1 + cot2A.

Solution:

LHS = (cos A – sin A + 1)/ (cos A + sin A – 1)

Dividing the numerator and denominator by sin A, we get;

= (cot A – 1 + cosec A)/(cot A + 1 – cosec A)

Using the identity cosec2A = 1 + cot2A ⇒ cosec2A – cot2A = 1,

= [cot A – (cosec2A – cot2A) + cosec A]/ (cot A + 1 – cosec A)

= [(cosec A + cot A) – (cosec A – cot A)(cosec A + cot A)] / (cot A + 1 – cosec A)

= [(cosec A + cot A) (1 – cosec A + cot A)]/ (1 – cosec A + cot A)]

= cosec A + cot A

= RHS

Hence proved.

9. Prove that: (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

[Hint: Simplify LHS and RHS separately]

Solution:

LHS = (cosec A – sin A)(sec A – cos A)

= [(1/sin A) – sin A) [(1/cos A) – cos A]

= [(1 – sin2A)/ sin A] [(1 – cos2A)/ cos A]

Using the identity sin2A + cos2A = 1,

= (cos2A/sin A) (sin2A/cos A)

= cos A sin A….(i)

RHS = 1/(tan A + cot A)

= 1/[(sin A/cos A) + (cos A/sin A)]

= (sin A cos A)/ (sin2A + cos2A)

= (sin A cos A)/1

= sin A cos A….(ii)

From (i) and (ii),

LHS = RHS

i.e. (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

Hence proved.

10. If a sin θ + b cos θ = c, prove that a cosθ – b sinθ = √(a2 + b2 – c2).

Solution:

Given,

a sin θ + b cos θ = c

Squaring on both sides,

(a sin θ + b cos θ)2 = c2

a2 sin2θ + b2 cos2θ + 2ab sin θ cos θ = c2

Using the identity sin2A + cos2A = 1,

a2(1 – cos2θ) + b2(1 – sin2θ) + 2ab sin θ cos θ = c2

a2 – a2 cos2θ + b2 – b2 sin2θ + 2ab sin θ cos θ = c2

a2 + b2 – c2 = a2 cos2θ + b2 sin2θ – 2ab sin θ cos θ

a2 + b2 – c2 = (a cos θ – b sin θ )2

⇒ a cos θ – b sin θ = √(a2 + b2 – c2)

Hence proved.

## Maths Questions

Q.1.  300 मीटर की दूरी से क्षैffतिज तल पर स्थित किसी शीर्ष की चोटी का उन्नयन कोण 30 अंश है। शिखर की ऊंचाई ज्ञात करो –

[A] 145.20मीटर

[B] 173.20 मीटर

[C] 145मीटर

[D] 200मीटर

Q.2.  505 मीटर लंबा एक खम्भा दीवार के सहारे जमीन के 60 अंश के कोण पर टिका हुआ हो, तो खंभा दीवार की किस ऊंचाई पर पहुंचेगा –

[A] 20.36मीटर

[B] 12.99 मीटर

[C] 14मीटर

[D] 18 मीटर

Q.3. कितने वर्षो में 12 प्रतिशत वार्षिक दर से 3000 रूपए का साधारण ब्याज 1080 रूपए हो जाएगा-

[A] 9वर्ष

[B] 4वर्ष

[C] 3 वर्ष

[D] 7 वर्ष

Q.4.  किसी त्रिभुज की भुजाएं 3:4:6 के अनुपात में है, त्रिभुज है –

[A] समकोण

[B] न्यूनकोण

[C] अधिक कोण

[D] इनमें से कोई नहीं

Q.5. किसी त्रिभुज की भुजाएं 3:4:5 के अनुपात में है। त्रिभुज के सबसे बड़े कोण का माप होगा –

[A] 63 अंश

[B] 68 अंश

[C] 71अंश

[D] 75 अंश

Q.6 एक शंकु, एक गोलाई, और एक बेलन एकसमान आधार वाले हैं, और उनकी ऊंचाई भी एक समान है। तद्नुसार उ न तीनों के आयतन का अनुपात क्या होगा –

[A]  3:3:1

[B]  3:7:1

[C]  1:2:3

[D] इनमें से कोई नहीं

Q.7.  यदि किसी वस्तु के क्रय मूल्य तथा विक्रय मूल्य 10:11 के अनुपात में हो तो लाभ प्रतिशत ज्ञात कीजिए-
[A] 10
[B] 6
[C] 5
[D] 1

Q.8. प्रथम 100 धन पूर्णाकों का औसत होगा –

[A] 49.5

[B] 47

[C] 52

[D] 50.5

Q.9. यदि किसी वस्तु को 200 प्रतिशत के लाभ से बेचा जाता है, तो इसके क्रय मूल्य का इसके विक्रय मूल्य से अनुपात होगा –

[A] 1:7

[B] 2:1

[C] 1:3

[D] 6:1

Q.10.  450 रूपए में कुछ टेनिस बाल खरीदे गए। यदि प्रत्येक बाल 15 रूपए सस्ता होता  तो उतनी ही राशि में 5 बाल और खरीदे जा सकते थे। खरीदे गए बाल की संख्या ज्ञात कीजिये –
[A] 6
[B] 7

[C] 10

[D] 74

Q.11. A sum of Rs. 25000 amounts to Rs. 31000 in 4 years at the rate of simple interest what is the rate of interest?

(a) 3%

(b) 4%

(c) 5 %

(d) 6 %

(e) None of these

Q.12. Kamla took a loan of Rs. 2400 with simple interest for as many years as the rate of interest. If she paid Rs. 864 as interest at the end of the loan period, what was the rate of interest?

(a) 3.6

(b) 6

(c) 18

(d) cannot determined

(e) None of these

Q.13. What is the present worth of Rs. 264 dues in 4 years at 10% simple interest per annum?

(a) 170.20

(b) 166

(c) 188.57

(d) 175.28

Q.14. A sum fetched a total simple interest of Rs. 8016.25 at the rate of 6 p.c.p.a in 5 years what is the sum?

(a) 24720.83

(b) 26730.33

(c) 26720.83

(d) 26710.63

(e) None of these

Q.15.  8. Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will Rs 800 become in 3 years?

(a) 1020.80

(b) 1025

(c) 1052

Q. 16.  The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is

a.   74

b.   78

c.   80

d.   90

Q.17.   The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a.   3

b.   6

c.   9

d.   8.2

Q.18.  √6+√6+√6+… is equal to –

a. 2

b. 5
c. 4
d. 3

Q.19. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Q.20. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67