All Trigonometry Math Notes PDF In Hindi {For All Competitive Exams}

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Trigonometry Questions And Answers

Question 1: Prove the following identity:

Solution:

LHS = (secθ – coseθ)(1 + tanθ + cotθ)

= (sinθcosθsinθcosθsinθ−cosθsinθcosθ)(sinθcosθ+1sinθcosθsinθcosθ+1sinθcosθ)

[Using identities, secθ = 1cosθ1cosθ, coseθ = 1sinθ1sinθ, tanθ = sinθcosθsinθcosθ and cotθ = cosθsinθcosθsinθ]

= (sinθcosθ)(sinθ cosθ+1)sin2θ cos2θ(sinθ−cosθ)(sinθ cosθ+1)sin2θ cos2θ

= (sinθcosθ)(sinθ cosθ+sin2θ+cos2θ)sin2θ cos2θ(sinθ−cosθ)(sinθ cosθ+sin2θ+cos2θ)sin2θ cos2θ

[Using, sin2θ+cos2θsin2θ+cos2θ = 1]


= tanθ sec θ – cotθcosecθ

= RHS

Question 2:Prove the following identity:

Solution:

LHS = 1 + 2sec2A tan2A – sec4A – tan4A

= 1 – (sec4A – 2sec2A tan2A + tan4A)

[Using identity, sec2A – tan2A = 1]

= 1 – 1

= 0

= RHS

Question 3: Prove the following identity:

Solution:

LHS = [$\frac{1 – sinA cosA}{cosA(secA – cosecA)}][][\frac{ sin^2 A – cos^2A}{sin^3A + cos^3A}]=[]=[\frac{1 – sinA cosA}{cosA(\frac{1}{cosA} – \frac{1}{sinA})}][][\frac{(sin A + cos A)(sin A – cos A)}{ (sin A + cos A)(sin^2 A – sin A cos A + cos^2A)}$]

[Using identities, secθ = 1cosθ1cosθ, coseθ = 1sinθ1sinθ, sin2A – cos2A = (sin A + cos A)(sin A – cos A) and sin3A + cos3A = (sin A + cos A)(sin2 A – sin A cos A + cos2A)]
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[1sinAcosAsinAcosAsinA1−sinAcosAsinA−cosAsinA][sinAcosAsin2AsinAcosA+cos2AsinA−cosAsin2A−sinAcosA+cos2A] [By cancelling common terms]

= sin A[1sinAcosAsinAcosA1−sinAcosAsinA−cosA] * [sinAcosA1sinAcosAsinA−cosA1−sinAcosA] [Using identity sin2 A + cos2 A = 1]

= sin A (By cancelling common terms)

= RHS

Trigonometry Test Questions

Lets us solve some more trigonometric examples using their identities:

Solved Examples

Question 1: Prove the following identity:


Solution:

LHS = (secθ – 1)2 – (tanθ – sinθ)2

= (1cosθ1cosθ – 1)2 – (sinθcosθsinθcosθ – sinθ)2

= (1cosθcosθ1−cosθcosθ)2 – sin2θcos2θsin2θcos2θ(1 – sinθ * cosθsinθcosθsinθ)2

= (1cosθcosθ1−cosθcosθ)2 – (1 – cosθ)2 sin2θcos2θsin2θcos2θ

= (1 – cosθ)2 [1cos2θsin2θcos2θ1cos2θ−sin2θcos2θ]

Question 2:Prove the following identity:

Solution:

LHS = tan3θ1+tan2θtan3θ1+tan2θ + cot3θ1+cot2θcot3θ1+cot2θ

tan3θsec2θtan3θsec2θ + cot3θcose2θcot3θcose2θ

Trigonometry Sample Questions

Solved Examples

Question 1:

Solution:

If tan A + sin A = m ……………… (1)

tan A – sin A = n ………………. (2)

Step 1:
Adding (1) and (2)

Step 2:
Subtracting (2) from (1)

Now

[Using, (a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab]

Question 2: 
Solution:

12+112+1 * 21212−12−1 cos θ

= (2‾√2 – 1)cos θ

Question 3:If x sin3θ+ y cos3θ = sin θ cos θ and x sin θ – y cos θ = 0. Prove that x2 + y2 = 1.

Solution:

x sin3θ + y cos3θ = sin θ cos θ ………………… (i)

x sin θ – y cos θ= 0 …………………. (ii)

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from (i) and (iii)

y = sinθ ……………… (iv)

From (iii) and (iv), x = cosθ

=> x2 + y2 = sin2θ + cos2θ

Question 4: 
Solution:

[using identity 1 – cos 2A = 2sin2 A similarly, 1 – cos 8A = 2sin24A and 1 – cos 4A = 2sin22A ]

[Using identity, sin2A = 2sin A cos A]

Question 5: 
Solution:

Step 1:

[Using identity, cos 3A = 4cos3A – 3cosA]

[Using, cos2A = 1+cos2A21+cos2A2]

or

Step 2:
let A = 15

About Trigonometry PDF Notes

These are the below topics have been covered in this PDF. We’re sure you all will like this.

  • Trigonometric Equation And Identities
  • Relation Between Trigonometric Ratios  
  • Ratio Description
  • Make New Relation Of Trigonometry
  • Identities Trigonometric Equation
  • Trigonometric Ratios Of Complementary Angles
  • Trigonometry Formulas

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