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

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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)]

[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:

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 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

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