Hello friend!! NCERT Solutions me Aaj hum Class 10th Math Chapter 8 Introduction to Trigonometry (Triokonmitiy ka parichay) ke exercise 8.2 ke sabhi question ko solve karne jaa rahe hai. Agar aap iss exercise ke question ko solve karna sikhna chahte hai to ye article aapke liye bahut helpful hoga.

Iss exercise me total 4 questions hai. Jisme Question - 1 me (i) to (v) questions, Question - 2 me 4 Multi choice questions, Question - 3 details solution question aur Question - 4 true & false Question hai. Inn sabhi Question ko aaj hum full details ke sath solve karne jaa rahe hai.

Agar aapko iss exercise ke kisi bhi question me koi bhi doubt yaa problem ho toh aap hume comment karke puchh sakte hai. Toh chaliye bina koi deri kiye Aaj ka lesson start karte hai.

NCERT Solutions Class 10 Math Exercise 8.2 Introduction to Trigonometry Hindi

Exercise 8.2

(1). Nimnlikhit ke maan nikaliye :

(i) sin 60° cos 30° + sin 30° cos 60°

(ii) 2 tan²45° + cos²30° - sin²60°

(iii) `\frac{cos 45°}{sec 30° + cosec 30°}`

(iv) `\frac{sin 30° + tan 45° - cosec 60°}{sec 30° + cos 60° + cot 45°}`

(v) `\frac{5 cos^2 60° + 4 sec^2 30° - tan^2 45°}{sin^2 30° + cos^2 30°}`

Solution-(i). sin 60° cos 30° + sin 30° cos 60°

`=\frac{√3}2\times\frac{√3}2+\frac{1}2\times\frac{1}2`

`=\frac{3}4+\frac{1}4`

`=\frac{3+1}4`

`=\frac{4}4`

`= 1` Ans.

Solution - (ii). 2 tan²45° + cos²30° - sin²60°

2 tan²45° + cos²30° - sin²60°

= 2×(1)²+ (√3/2)²- (√3/2)²

= 2×1

= 2 Ans.

Solution - (iii).

`\frac{cos 45°}{sec 30° + cosec 30°} = \frac{\frac1sqrt2}{\frac2sqrt3 + 2}`

`= \frac{\frac1sqrt2}{\frac{2+2sqrt3}sqrt3}`

`= \frac1sqrt2times\frac{sqrt3}{2(1 + sqrt3)}`

`= \frac{1}{2sqrt2}times\frac{sqrt3}{(1 + sqrt3)}`

Ansh (Numerator) aur Har (Denominator) me (1-√3) se multiply karne par,

`= \frac{1}{2sqrt2}times\frac{sqrt3×(1-sqrt3)}{(1+sqrt3)(1-sqrt3)}`

`= \frac{1}{2sqrt2}times\frac{sqrt3×(1-sqrt3)}{(1)^2 - (sqrt3)^2}`

`= \frac{1}{2sqrt2}times\frac{sqrt3×(1-sqrt3)}{1 - 3}`

`= \frac{1}{2sqrt2}times\frac{sqrt3×(1-sqrt3)}{-2}`

`= \frac{sqrt3×(1-sqrt3)}{-4sqrt2}`

`= \frac{sqrt3×(sqrt3-1)}{4sqrt2}`

Phir se Ansh (Numerator) aur Har (Denominator) me √2 se multiply karne par,

`= \frac{sqrt3×sqrt2×(sqrt3-1)}{4sqrt2×sqrt2}`

`= \frac{3sqrt2-sqrt6}8` Ans.

Solution - (iv) `\frac{sin 30° + tan 45° - cosec 60°}{sec 30° + cos 60° + cot 45°}`

`= \frac{\frac{1}2+1-\frac2sqrt3}{\frac{2}sqrt3+\frac{1}2+1`

`= \frac{\frac{sqrt3+2sqrt3-4}{2sqrt3}}{\frac{4+sqrt3+2sqrt3}{2sqrt3}`

`= \frac{3sqrt3-4}{3sqrt3+4}`

Ansh (Numerator) aur Har (Denominator) me `(3sqrt3-4)` se multiply karne par,

`= \frac{(3sqrt3-4)(3sqrt3-4)}{(3sqrt3+4)(3sqrt3-4)}`

`= \frac{(3sqrt3-4)^2}{(3sqrt3)^2-4^2}`

`= \frac{(3sqrt3)^2+4^2-2×3sqrt3×4}{27-16}`

`= \frac{27+16-24sqrt3}11`

`= \frac{43-24sqrt3}11` Ans.

Solution - (v) `\frac{5 cos^2 60° + 4 sec^2 30° - tan^2 45°}{sin^2 30° + cos^2 30°}`

`=\frac{5times\left(\frac{1}2\right)^2 + 4times\left(\frac2{\sqrt3}\right)^2 - (1)^2}{(\frac{1}2\right)^2 + (\frac{\sqrt3}2\right)^2 }`

`=\frac{5times\left(\frac{1}4) + 4times\left(\frac4{\3}) - 1}{\frac{1}4 + \frac{3}4 }`

`=\frac{\frac{5}4 + \frac{16}3 - 1}{\frac{1}4 + \frac{3}4 }`

`=\frac{\frac{15+64-12}12}{\frac{1+3}4}`

`=\frac{\frac{67}12}{\frac{4}4}`

`=\frac{67}12\times\1`

`=\frac{67}12` Ans.

NCERT Solutions math Chapter 8 Introduction to Trigonometry Ex 8.2

2. Sahi vikalp chuniye aur apne vikalp ka auchity dijiye:

(i)`\frac{2 tan 30°}{1 + tan^2 30°} =`

(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°

Solution -: `\frac{2 tan 30°}{1 + tan^2 30°} =\frac{2times\frac{1}sqrt3}{1 + (\frac1{\sqrt3}\right)^2 }`

`=\frac{\frac{2}sqrt3}{1 + \frac{1}3 }`

`=\frac{\frac{2}sqrt3}{\frac{3+1}3 }`

`=\frac{\frac{2}sqrt3}{\frac{4}3 }`

`=\frac{2}sqrt3times\frac{3}4`

`=\frac{sqrt3}2`

`= sin 60°`

Correption - (A) sin 60°✅

(ii)`\frac{1 - tan^2 45°}{1 + tan^2 45°}`

(A) tan 90° (B) 1 (C) sin 45° (D) 0

Solution -: (ii) `\frac{1 - tan^2 45°}{1 + tan^2 45°}`

`=\frac{1 - (1)^2}{1 + (1)^2}`

`=\frac{0}{2}`

`= 0`

Correct option - (D) 0 ✅

(iii) sin 2A = 2 sin A tab saty hota hai, jabki A barabar hai:

(A) 0° (B) 30° (C) sin 45° (D) 60°

Solution -: (iii) sin 2A = 2 sin A me A = 0° rakhne par,

L.H.S. = sin 2A

= sin 2×0

= sin 0°

= 0

R.H.S. = 2 sin A

= 2 sin 0°

= 2×0

= 0

L.H.S. = R.H.S.

Correct option - (A) 0° ✅

(iv)`\frac{2 tan 30°}{1 - tan^2 30°} ` barabar hai:

(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

Solution -(iv).`\frac{2 tan 30°}{1 - tan^2 30°} `

`=\frac{2times\frac{1}sqrt3}{1 - (\frac1{\sqrt3}\right)^2 }`

`=\frac{\frac{2}sqrt3}{1 - \frac{1}3 }`

`=\frac{\frac{2}sqrt3}{\frac{3-1}3 }`

`=\frac{\frac{2}sqrt3}{\frac{2}3 }`

`=\frac{2}sqrt3times\frac{3}2`

`=sqrt3`

`= tan 60°`

Correption - (C) tan 60°✅

NCERT solutions Class 10th Math Exercise 8.2 Question - 3 and Question - 4

(3). Yadi tan (A+B) =√3 aur tan (A-B) = 1/√3 ; 0°<A+B≤90° ; A>B toh A aur B ka maan gyaat kijiye.

Solution -: tan (A+B) = √3

⇒ tan (A+B) = tan 60°

⇒ (A+B) = 60° ....................(i)

Aur tan (A-B) = 1/√3

⇒ tan (A-B) = tan 30°

⇒ (A-B) = 30° ....................(ii)

Equation (i) ko (ii) se Add (jodane) par,

⇒ 2A = 90°

⇒ A = 45°

Phir A = 45° equation (i) me rakhne par,

A+B = 60°

45°+B = 60

B = 60°- 45°

= 15°

So, A = 45° and B = 15° Ans.

4. Batayiye ki nimnlikhit me kaun - kaun Saty (True) hain yaa Asaty (False) hain. Karan sahit apne uttar ki pushti kijiye.

(i) sin (A+B) = sin A + sin B

(ii) θ me vridhi hone ke saath sin θ ke maan me bhi vridhi hoti hai.

(iii) θ me vridhi hone ke saath cos θ ke maan me bhi vridhi hoti hai.

(iv) θ ke sabhi maano par sin θ = cos θ

(v) A = 0° par cot A paribhashit nahi hai.

(i) sin (A+B) = sin A + sin B [ False ]

Karan-: Maana A = 30° aur B = 60° L.H.S. me rakhne par,

L.H.S.= sin (A+B)

= sin (30°+60°)

= sin 90°

= 1

R.H.S. = sin A + sin B

= sin 30° + sin 60°

= 1/2 + √3/2

= (1+√3)/2

So, L.H.S. ≠ R.H.S.

Isiliye, yah Asaty hai.

(ii) θ me vridhi hone ke saath sin θ ke maan me bhi vridhi hoti hai. [ True ]

Karan-: Jab θ = 0° , tab sin 0° = 0

Jab θ = 30°, tab sin 30° = 1/2

Jab θ = 45°, tab sin 45° = 1/√2

Jab θ = 60°, tab sin 60° = √3/2

Jab θ = 90°, tab sin 90° = 1

Hum dekh sakte hai ki, jaise - jaise θ me vridhi hoti hai waise - waise sin θ me bhi vridhi hoti hai. Isiliye, yah Saty hai

(iii) θ me vridhi hone ke saath cos θ ke maan me bhi vridhi hoti hai. [ False ]

Karan-: Jab θ = 0° , tab cos 0° = 1

Jab θ = 30°, tab cos 30° = √3/2

Jab θ = 45°, tab cos 45° = 1/√2

Jab θ = 60°, tab cos 60° = 1/2

Jab θ = 90°, tab cos 90° = 0

Hum dekh sakte hai ki, jaise - jaise θ me vridhi hoti hai waise - waise cos θ me kami hoti hai. Isiliye ye Asaty hai.

(iv) θ ke sabhi maano par sin θ = cos θ [False]

Karan-: Maan θ = 30° , tab sin 30° = 1/2 aur cos 30° = √3/2

Toh, sin 30° ≠ cos 30°

Isiliye, yah Asaty hai.

(v) A = 0° par cot A paribhashit nahi hai. [ True ]

Hum jaante hai ki, cot A = cos A/sin A

Ab A = 0° maan rakhne par, cot 0° = sin 0°/cos 0°

= 1/0

= Aparibhashit

∴ A = 0° par cot A paribhashit nahi hai. Isiliye, yah Saty hai.

Ye bhi shikhiye -:

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