Mathematics: Trignometry Trignometricratio Mock Test – Free Online Practice

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Practice Questions (20 of 20)

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1. If 3 sec²x - 4 = 0, then the value of x (0 < x < 90°) will be :
   यदि 3 sec²x - 4 = 0 है, तो x (0 < x < 90°) का मान क्या होगा?
   (A) (a) 15°

   (B) (b) 45°

   (C) (c) 30°

   (D) (d) 60°
   ✅ Answer & Explanation

   Sahi jawab: C) (c) 30°

   Explanation: Method 1 (Direct Trigonometric Table Method): Step 1: Given equation ko simplify karte hain: $$3\sec^2 x = 4 \implies \sec^2 x = \frac{4}{3}$$ Step 2: Square root lene par ($0 < x < 90^\circ$ acute angle hai): $$\sec x = \frac{2}{\sqrt{3}}$$ Step 3: Standard table se angle check karein: $$\sec 30^\circ = \frac{2}{\sqrt{3}} \implies x = 30^\circ$$ Correct option (c) 30° hai. Method 2 (Cosine Inversion Method): $$\cos^2 x = \frac{1}{\sec^2 x} = \frac{3}{4} \implies \cos x = \frac{\sqrt{3}}{2}$$ $$\cos 30^\circ = \frac{\sqrt{3}}{2} \implies x = 30^\circ$$ Sahi uttar (c) hai.
2. If r sin θ = 1, r cos θ = √3, then the value of (√3 tan θ + 1) is :
   यदि r sin θ = 1, r cos θ = √3 है, तो (√3 tan θ + 1) का मान क्या होगा?
   (A) (a) √3

   (B) (b) 1/√3

   (C) (c) 1

   (D) (d) 2
   ✅ Answer & Explanation

   Sahi jawab: D) (d) 2

   Explanation: Method 1 (Ratio Division Method): Step 1: Dono equations ko divide karke $r$ ko eliminate karte hain: $$\frac{r\sin\theta}{r\cos\theta} = \frac{1}{\sqrt{3}} \implies \tan\theta = \frac{1}{\sqrt{3}}$$ Step 2: Ab target expression $(\sqrt{3}\tan\theta + 1)$ me $\tan\theta$ ki value put karte hain: $$\sqrt{3}\left(\frac{1}{\sqrt{3}}\right) + 1 = 1 + 1 = 2$$ Correct option (d) 2 hai. Method 2 (Angle Determination Method): $$\tan\theta = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ$$ $$\sqrt{3}\tan 30^\circ + 1 = \sqrt{3}\left(\frac{1}{\sqrt{3}}\right) + 1 = 2$$ Sahi uttar (d) hai.
3. 2 tan 30° / (1 - tan² 30°) = ?
   2 tan 30° / (1 - tan² 30°) = ?
   (A) (a) 3

   (B) (b) 1/3

   (C) (c) √3

   (D) (d) 1/√3
   ✅ Answer & Explanation

   Sahi jawab: C) (c) √3

   Explanation: Method 1 (Double Angle Formula Shortcut Method): Step 1: Standard formula yaad rakhein: $$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$$ Step 2: Yahan $A = 30^\circ$ diya gaya hai: $$\frac{2\tan 30^\circ}{1 - \tan^2 30^\circ} = \tan(2 \times 30^\circ) = \tan 60^\circ = \sqrt{3}$$ Bina value calculate kiye direct formula se 2 second me answer aa gaya. Correct option (c) √3 hai. Method 2 (Value Substitution Method): $\tan 30^\circ = \frac{1}{\sqrt{3}}$ substitute karein: $$\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1 - \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3}$$ Sahi uttar (c) hai.
4. If x sin 60° · tan 30° = sec 60° · cot 45°, then the value of x is :
   यदि x sin 60° · tan 30° = sec 60° · cot 45° है, तो x का मान क्या होगा?
   (A) (a) √3

   (B) (b) 1/√3

   (C) (c) 4

   (D) (d) 4√3
   ✅ Answer & Explanation

   Sahi jawab: C) (c) 4

   Explanation: Method 1 (Standard Table Value Method): Step 1: Sabhi standard angles ki values likhte hain: - $\sin 60^\circ = \frac{\sqrt{3}}{2}$ - $\tan 30^\circ = \frac{1}{\sqrt{3}}$ - $\sec 60^\circ = 2$ - $\cot 45^\circ = 1$ Step 2: Equation me values substitute karein: $$x \times \left(\frac{\sqrt{3}}{2}\right) \times \left(\frac{1}{\sqrt{3}}\right) = 2 \times 1$$ Step 3: Solve karein: $$x \times \frac{1}{2} = 2 \implies x = 4$$ Correct option (c) 4 hai. Method 2 (Product Simplification Method): $$\sin 60^\circ \times \tan 30^\circ = \sin 60^\circ \times \frac{\sin 30^\circ}{\cos 30^\circ} = \frac{\sqrt{3}}{2} \times \frac{1/2}{\sqrt{3}/2} = \frac{1}{2}$$ $$x \left(\frac{1}{2}\right) = 2 \implies x = 4$$ Sahi uttar (c) hai.
5. If x sin 60° · tan 30° - tan² 45° = cosec 60° · cot 30° - sec² 45°, then the value of x is :
   यदि x sin 60° · tan 30° - tan² 45° = cosec 60° · cot 30° - sec² 45° है, तो x का मान क्या होगा?
   (A) (a) 2

   (B) (b) - 2

   (C) (c) 6

   (D) (d) - 4
   ✅ Answer & Explanation

   Sahi jawab: A) (a) 2

   Explanation: Method 1 (Step-by-Step Value Substitution Method): Step 1: LHS evaluate karte hain: $$\text{LHS} = x \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{\sqrt{3}}\right) - (1)^2 = \frac{x}{2} - 1$$ Step 2: RHS evaluate karte hain: $$\csc 60^\circ = \frac{2}{\sqrt{3}}, \quad \cot 30^\circ = \sqrt{3}, \quad \sec 45^\circ = \sqrt{2}$$ $$\text{RHS} = \left(\frac{2}{\sqrt{3}}\right) (\sqrt{3}) - (\sqrt{2})^2 = 2 - 2 = 0$$ Step 3: LHS = RHS equate karte hain: $$\frac{x}{2} - 1 = 0 \implies \frac{x}{2} = 1 \implies x = 2$$ Correct option (a) 2 hai. Method 2 (Direct Simplification): $$\frac{x}{2} - 1 = 2 - 2 = 0 \implies x = 2$$ Sahi uttar (a) hai.
6. The numerical value of (cos² 45° / sin² 60°) + (cos² 60° / sin² 45°) - (tan² 30° / cot² 45°) - (sin² 30° / cot² 30°) is :
   (cos² 45° / sin² 60°) + (cos² 60° / sin² 45°) - (tan² 30° / cot² 45°) - (sin² 30° / cot² 30°) का संख्यात्मक मान क्या है?
   (A) (a) 1 (1/4)

   (B) (b) 3/4

   (C) (c) 1/4

   (D) (d) 1/2
   ✅ Answer & Explanation

   Sahi jawab: B) (b) 3/4

   Explanation: Method 1 (Term-by-Term Evaluation Method): Step 1: Chaaron terms ko alag-alag solve karte hain: - Term 1: $\frac{\cos^2 45^\circ}{\sin^2 60^\circ} = \frac{(1/\sqrt{2})^2}{(\sqrt{3}/2)^2} = \frac{1/2}{3/4} = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}$ - Term 2: $\frac{\cos^2 60^\circ}{\sin^2 45^\circ} = \frac{(1/2)^2}{(1/\sqrt{2})^2} = \frac{1/4}{1/2} = \frac{1}{2}$ - Term 3: $\frac{\tan^2 30^\circ}{\cot^2 45^\circ} = \frac{(1/\sqrt{3})^2}{(1)^2} = \frac{1/3}{1} = \frac{1}{3}$ - Term 4: $\frac{\sin^2 30^\circ}{\cot^2 30^\circ} = \frac{(1/2)^2}{(\sqrt{3})^2} = \frac{1/4}{3} = \frac{1}{12}$ Step 2: Sabhi values ko add/subtract karte hain: $$\text{Value} = \frac{2}{3} + \frac{1}{2} - \frac{1}{3} - \frac{1}{12}$$ Step 3: LCM = 12 lekar calculate karein: $$\text{Value} = \left(\frac{2}{3} - \frac{1}{3}\right) + \frac{1}{2} - \frac{1}{12} = \frac{1}{3} + \frac{1}{2} - \frac{1}{12}$$ $$\frac{4 + 6 - 1}{12} = \frac{9}{12} = \frac{3}{4}$$ Correct option (b) 3/4 hai. Method 2 (Common Denominator Method): $$\frac{8 + 6 - 4 - 1}{12} = \frac{9}{12} = \frac{3}{4}$$ Sahi uttar (b) hai.
7. The value of sin² 30° cos² 45° + 4 tan² 30° + (1/2) sin² 90° + 2 cos 90° is :
   sin² 30° cos² 45° + 4 tan² 30° + (1/2) sin² 90° + 2 cos 90° का मान क्या होगा?
   (A) (a) 15/8

   (B) (b) 47/24

   (C) (c) 23/12

   (D) (d) 2
   ✅ Answer & Explanation

   Sahi jawab: B) (b) 47/24

   Explanation: Method 1 (Direct Table Substitution Method): Step 1: Sabhi standard angle values rakhein: - $\sin 30^\circ = \frac{1}{2} \implies \sin^2 30^\circ = \frac{1}{4}$ - $\cos 45^\circ = \frac{1}{\sqrt{2}} \implies \cos^2 45^\circ = \frac{1}{2}$ - $\tan 30^\circ = \frac{1}{\sqrt{3}} \implies \tan^2 30^\circ = \frac{1}{3}$ - $\sin 90^\circ = 1 \implies \sin^2 90^\circ = 1$ - $\cos 90^\circ = 0$ Step 2: Har term calculate karein: - Term 1: $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$ - Term 2: $4 \times \frac{1}{3} = \frac{4}{3}$ - Term 3: $\frac{1}{2} \times 1 = \frac{1}{2}$ - Term 4: $2 \times 0 = 0$ Step 3: Inhe add karein (LCM = 24): $$\text{Total} = \frac{1}{8} + \frac{4}{3} + \frac{1}{2} = \frac{3 + 32 + 12}{24} = \frac{47}{24}$$ Correct option (b) 47/24 hai. Method 2 (Group Addition Method): $$\left(\frac{1}{8} + \frac{1}{2}\right) + \frac{4}{3} = \frac{5}{8} + \frac{4}{3} = \frac{15 + 32}{24} = \frac{47}{24}$$ Sahi uttar (b) hai.
8. The value of [4 tan² 30° + (1/4) sin² 90° + (1/8) cot² 60° + sin² 30° · cos² 45°] / [sin 60° cos 30° - cos 60° sin 30°] is :
   [4 tan² 30° + (1/4) sin² 90° + (1/8) cot² 60° + sin² 30° · cos² 45°] / [sin 60° cos 30° - cos 60° sin 30°] का मान क्या होगा?
   (A) (a) 1 (3/4)

   (B) (b) 4

   (C) (c) 2 (1/2)

   (D) (d) 3 (1/2)
   ✅ Answer & Explanation

   Sahi jawab: D) (d) 3 (1/2)

   Explanation: Method 1 (Compound Angle & Table Method): Step 1: Denominator me identity $\sin(A - B) = \sin A\cos B - \cos A\sin B$ lagate hain: $$\text{Denominator} = \sin(60^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$$ Step 2: Numerator evaluate karte hain: - $4\tan^2 30^\circ = 4\left(\frac{1}{3}\right) = \frac{4}{3}$ - $\frac{1}{4}\sin^2 90^\circ = \frac{1}{4}(1) = \frac{1}{4}$ - $\frac{1}{8}\cot^2 60^\circ = \frac{1}{8}\left(\frac{1}{3}\right) = \frac{1}{24}$ - $\sin^2 30^\circ\cos^2 45^\circ = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$ Step 3: Numerator ka sum calculate karein (LCM = 24): $$\text{Num} = \frac{32 + 6 + 1 + 3}{24} = \frac{42}{24} = \frac{7}{4}$$ Step 4: Numerator ko Denominator se divide karein: $$\text{Value} = \frac{\frac{7}{4}}{\frac{1}{2}} = \frac{7}{4} \times 2 = \frac{7}{2} = 3\frac{1}{2}$$ Correct option (d) 3 (1/2) hai. Method 2 (Direct Simplification): $$\frac{7/4}{1/2} = \frac{7}{2} = 3.5$$ Sahi uttar (d) hai.
9. The value of (4/3) tan² 60° + 3 cos² 30° - 2 sec² 30° - (3/4) cot² 60° is equal to :
   (4/3) tan² 60° + 3 cos² 30° - 2 sec² 30° - (3/4) cot² 60° किसके बराबर है?
   (A) (a) 8/3

   (B) (b) 5/4

   (C) (c) 10/3

   (D) (d) 7/3
   ✅ Answer & Explanation

   Sahi jawab: C) (c) 10/3

   Explanation: Method 1 (Term-by-Term Substitution Method): Step 1: Sabhi standard values rakhein: - $\tan 60^\circ = \sqrt{3} \implies \tan^2 60^\circ = 3$ - $\cos 30^\circ = \frac{\sqrt{3}}{2} \implies \cos^2 30^\circ = \frac{3}{4}$ - $\sec 30^\circ = \frac{2}{\sqrt{3}} \implies \sec^2 30^\circ = \frac{4}{3}$ - $\cot 60^\circ = \frac{1}{\sqrt{3}} \implies \cot^2 60^\circ = \frac{1}{3}$ Step 2: Har term ko multiply karein: - Term 1: $\frac{4}{3} \times 3 = 4$ - Term 2: $3 \times \frac{3}{4} = \frac{9}{4}$ - Term 3: $2 \times \frac{4}{3} = \frac{8}{3}$ - Term 4: $\frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$ Step 3: Terms ko arrange karke solve karein: $$\text{Value} = 4 + \left(\frac{9}{4} - \frac{1}{4}\right) - \frac{8}{3} = 4 + \frac{8}{4} - \frac{8}{3}$$ $$\text{Value} = 4 + 2 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{18 - 8}{3} = \frac{10}{3}$$ Correct option (c) 10/3 hai. Method 2 (LCM of Fractions Method): $$4 + \frac{9}{4} - \frac{8}{3} - \frac{1}{4} = 4 + 2 - \frac{8}{3} = \frac{10}{3}$$ Sahi uttar (c) hai.
10. The value of (tan 30° · cosec 60° + tan 60° · sec 30°) / (sin² 30° + 4 cot² 45° - sec² 60°) is :
    (tan 30° · cosec 60° + tan 60° · sec 30°) / (sin² 30° + 4 cot² 45° - sec² 60°) का मान क्या होगा?
    (A) (a) 2/3

    (B) (b) 32/3

    (C) (c) 8/3

    (D) (d) 32/99
    ✅ Answer & Explanation

    Sahi jawab: B) (b) 32/3

    Explanation: Method 1 (Standard Angles Table Method): Step 1: Numerator evaluate karte hain: $$\tan 30^\circ = \frac{1}{\sqrt{3}}, \quad \csc 60^\circ = \frac{2}{\sqrt{3}}, \quad \tan 60^\circ = \sqrt{3}, \quad \sec 30^\circ = \frac{2}{\sqrt{3}}$$ $$\text{Numerator} = \left(\frac{1}{\sqrt{3}} \times \frac{2}{\sqrt{3}}\right) + \left(\sqrt{3} \times \frac{2}{\sqrt{3}}\right) = \frac{2}{3} + 2 = \frac{2 + 6}{3} = \frac{8}{3}$$ Step 2: Denominator evaluate karte hain: $$\sin 30^\circ = \frac{1}{2}, \quad \cot 45^\circ = 1, \quad \sec 60^\circ = 2$$ $$\text{Denominator} = \left(\frac{1}{2}\right)^2 + 4(1)^2 - (2)^2 = \frac{1}{4} + 4 - 4 = \frac{1}{4}$$ Step 3: Numerator ko Denominator se divide karein: $$\text{Value} = \frac{\frac{8}{3}}{\frac{1}{4}} = \frac{8}{3} \times 4 = \frac{32}{3}$$ Correct option (b) 32/3 hai. Method 2 (Direct Fraction Inversion): $$\frac{8/3}{1/4} = \frac{8 \times 4}{3} = \frac{32}{3}$$ Sahi uttar (b) hai.
11. The value of [cosec²30° sin²45° + sec²60°] / [tan60° cosec²45° - sec²60° tan45°] is :
    [cosec²30° sin²45° + sec²60°] / [tan60° cosec²45° - sec²60° tan45°] का मान क्या होगा?
    (A) (a) 3(2 + √3)

    (B) (b) 2(√3 - 2)

    (C) (c) - 2√3 - 2

    (D) (d) - 3(2 + √3)
    ✅ Answer & Explanation

    Sahi jawab: D) (d) - 3(2 + √3)

    Explanation: Method 1 (Standard Values and Rationalization Method): Step 1: Numerator ke sabhi terms evaluate karte hain: - $\csc 30^\circ = 2 \implies \csc^2 30^\circ = 4$ - $\sin 45^\circ = \frac{1}{\sqrt{2}} \implies \sin^2 45^\circ = \frac{1}{2}$ - $\sec 60^\circ = 2 \implies \sec^2 60^\circ = 4$ $$\text{Numerator} = 4 \left(\frac{1}{2}\right) + 4 = 2 + 4 = 6$$ Step 2: Denominator ke terms evaluate karte hain: - $\tan 60^\circ = \sqrt{3}$ - $\csc 45^\circ = \sqrt{2} \implies \csc^2 45^\circ = 2$ - $\sec 60^\circ = 2 \implies \sec^2 60^\circ = 4$ - $\tan 45^\circ = 1$ $$\text{Denominator} = \sqrt{3}(2) - 4(1) = 2\sqrt{3} - 4 = 2(\sqrt{3} - 2)$$ Step 3: Fraction simplify karke rationalise karein: $$\frac{6}{2(\sqrt{3} - 2)} = \frac{3}{\sqrt{3} - 2}$$ Conjugate $(\sqrt{3} + 2)$ se multiply aur divide karein: $$= \frac{3(\sqrt{3} + 2)}{(\sqrt{3})^2 - (2)^2} = \frac{3(2 + \sqrt{3})}{3 - 4} = -3(2 + \sqrt{3})$$ Correct option (d) - 3(2 + √3) hai. Method 2 (Sign Factor Method): Denominator me se minus common nikalne par: $$\frac{3}{-(2 - \sqrt{3})} = -3(2 + \sqrt{3})$$ Sahi uttar (d) hai.
12. The value of [sin²30° + cos²60° + sec45° sin45°] / [sec60° + cosec30°] is :
    [sin²30° + cos²60° + sec45° sin45°] / [sec60° + cosec30°] का मान क्या होगा?
    (A) (a) 1/4

    (B) (b) - 1/4

    (C) (c) - 3/8

    (D) (d) 3/8
    ✅ Answer & Explanation

    Sahi jawab: D) (d) 3/8

    Explanation: Method 1 (Table Value Method): Step 1: Numerator ki value calculate karte hain: - $\sin 30^\circ = \frac{1}{2} \implies \sin^2 30^\circ = \frac{1}{4}$ - $\cos 60^\circ = \frac{1}{2} \implies \cos^2 60^\circ = \frac{1}{4}$ - $\sec 45^\circ \times \sin 45^\circ = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$ $$\text{Numerator} = \frac{1}{4} + \frac{1}{4} + 1 = \frac{1}{2} + 1 = \frac{3}{2}$$ Step 2: Denominator ki value calculate karte hain: - $\sec 60^\circ = 2$ - $\csc 30^\circ = 2$ $$\text{Denominator} = 2 + 2 = 4$$ Step 3: Dono ko divide karein: $$\text{Value} = \frac{\frac{3}{2}}{4} = \frac{3}{8}$$ Correct option (d) 3/8 hai. Method 2 (Direct Simplification): $$\frac{0.25 + 0.25 + 1}{2 + 2} = \frac{1.5}{4} = \frac{3}{8}$$ Sahi uttar (d) hai.
13. Which among the following is an irrational quantity?
    निम्नलिखित में से कौन-सी एक अपरिमेय (irrational) राशि है?
    (A) (a) tan 30° · tan 60°

    (B) (b) sin 30°

    (C) (c) tan 45°

    (D) (d) cos 30°
    ✅ Answer & Explanation

    Sahi jawab: D) (d) cos 30°

    Explanation: Method 1 (Evaluation of Options Method): Step 1: Sabhi options ki numeric values check karein: - (a) $\tan 30^\circ \cdot \tan 60^\circ = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1$ (Rational number hai) - (b) $\sin 30^\circ = \frac{1}{2}$ (Rational number hai) - (c) $\tan 45^\circ = 1$ (Rational number hai) - (d) $\cos 30^\circ = \frac{\sqrt{3}}{2}$ Step 2: $\sqrt{3}$ ek non-terminating, non-repeating decimal hone ke karan irrational (अपरिमेय) number hai, isliye $\frac{\sqrt{3}}{2}$ bhi ek irrational quantity hai. Correct option (d) cos 30° hai. Method 2 (Definition Method): Keval $\cos 30^\circ$ ke expression me $\sqrt{3}$ bachta hai jo square root se bahar nahi nikal sakta, baaki sabhi options me $\sqrt{3}$ cancel ho jata hai ya present hi nahi hai. Sahi uttar (d) hai.
14. If sin (θ + 30°) = 3/√12, then the value of θ (0 < θ < 90°) is :
    यदि sin (θ + 30°) = 3/√12 है, तो θ (0 < θ < 90°) का मान क्या होगा?
    (A) (a) 60°

    (B) (b) 15°

    (C) (c) 45°

    (D) (d) 30°
    ✅ Answer & Explanation

    Sahi jawab: D) (d) 30°

    Explanation: Method 1 (Surds Simplification Method): Step 1: RHS ko simplify karte hain: $$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$$ $$\text{RHS} = \frac{3}{\sqrt{12}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$$ Step 2: Standard sine values se compare karein: $$\sin(\theta + 30^\circ) = \frac{\sqrt{3}}{2} = \sin 60^\circ$$ Step 3: Angles compare karein: $$\theta + 30^\circ = 60^\circ \implies \theta = 60^\circ - 30^\circ = 30^\circ$$ Correct option (d) 30° hai. Method 2 (Rationalization Method): $$\frac{3}{\sqrt{12}} = \sqrt{\frac{9}{12}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$ $$\theta + 30^\circ = 60^\circ \implies \theta = 30^\circ$$ Sahi uttar (d) hai.
15. If sin θ = √3 cos θ, 0° < θ < 90°, then the value of 2 sin²θ + sec²θ + sin θ · sec θ + cosec θ is :
    यदि sin θ = √3 cos θ, 0° < θ < 90° है, तो 2 sin²θ + sec²θ + sin θ · sec θ + cosec θ का मान क्या होगा?
    (A) (a) (33 + 10√3)/6

    (B) (b) (19 + 10√3)/6

    (C) (c) (33 + 10√3)/3

    (D) (d) (19 + 10√3)/3
    ✅ Answer & Explanation

    Sahi jawab: A) (a) (33 + 10√3)/6

    Explanation: Method 1 (Angle Determination Method): Step 1: Given equation se $\theta$ find karte hain: $$\frac{\sin\theta}{\cos\theta} = \sqrt{3} \implies \tan\theta = \sqrt{3} \implies \theta = 60^\circ$$ Step 2: $\theta = 60^\circ$ ke sabhi required values nikalte hain: - $\sin 60^\circ = \frac{\sqrt{3}}{2} \implies \sin^2 60^\circ = \frac{3}{4}$ - $\sec 60^\circ = 2 \implies \sec^2 60^\circ = 4$ - $\sin 60^\circ \cdot \sec 60^\circ = \frac{\sqrt{3}}{2} \times 2 = \sqrt{3}$ - $\csc 60^\circ = \frac{2}{\sqrt{3}}$ Step 3: Sabhi terms ko add karein: $$\text{Value} = 2\left(\frac{3}{4}\right) + 4 + \sqrt{3} + \frac{2}{\sqrt{3}} = \frac{3}{2} + 4 + \frac{3 + 2}{\sqrt{3}} = \frac{11}{2} + \frac{5}{\sqrt{3}}$$ Step 4: Denominator ko rationalize karke combine karein: $$\frac{11}{2} + \frac{5\sqrt{3}}{3} = \frac{11 \times 3 + 2 \times 5\sqrt{3}}{6} = \frac{33 + 10\sqrt{3}}{6}$$ Correct option (a) (33 + 10√3)/6 hai. Method 2 (Direct LCM Calculation): $$\frac{11}{2} + \frac{5}{\sqrt{3}} = \frac{11\sqrt{3} + 10}{2\sqrt{3}} = \frac{(11\sqrt{3} + 10)\sqrt{3}}{6} = \frac{33 + 10\sqrt{3}}{6}$$ Sahi uttar (a) hai.
16. If 0° < θ < 90° and cos²θ = 3(cot²θ - cos²θ), then the value of (1/2 sec θ + sin θ)⁻¹ is :
    यदि 0° < θ < 90° और cos²θ = 3(cot²θ - cos²θ) है, तो (1/2 sec θ + sin θ)⁻¹ का मान क्या होगा?
    (A) (a) √3 + 2

    (B) (b) 2(2 - √3)

    (C) (c) 2(√3 - 1)

    (D) (d) √3 + 1
    ✅ Answer & Explanation

    Sahi jawab: B) (b) 2(2 - √3)

    Explanation: Method 1 (Angle Extraction & Rationalization Method): Step 1: Diye gaye equation ko solve karte hain: $$\cos^2\theta = 3\left(\frac{\cos^2\theta}{\sin^2\theta} - \cos^2\theta\right) = 3\cos^2\theta\left(\frac{1}{\sin^2\theta} - 1\right)$$ Chuki $0^\circ < \theta < 90^\circ$ hai, $\cos^2\theta \ne 0$, dono taraf $\cos^2\theta$ cancel ho jayega: $$1 = 3(\csc^2\theta - 1) = 3\cot^2\theta$$ $$\cot^2\theta = \frac{1}{3} \implies \cot\theta = \frac{1}{\sqrt{3}} \implies \theta = 60^\circ$$ Step 2: Target expression me $\theta = 60^\circ$ rakhein: $$\frac{1}{2}\sec 60^\circ + \sin 60^\circ = \frac{1}{2}(2) + \frac{\sqrt{3}}{2} = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$$ Step 3: Iska reciprocal (inverse) nikal kar rationalize karein: $$\left(\frac{2 + \sqrt{3}}{2}\right)^{-1} = \frac{2}{2 + \sqrt{3}} = \frac{2(2 - \sqrt{3})}{(2)^2 - (\sqrt{3})^2} = \frac{2(2 - \sqrt{3})}{4 - 3} = 2(2 - \sqrt{3})$$ Correct option (b) 2(2 - √3) hai. Method 2 (Direct Step Inverse): $$\frac{1}{1 + \frac{\sqrt{3}}{2}} = \frac{2}{2 + \sqrt{3}} = 2(2 - \sqrt{3})$$ Sahi uttar (b) hai.
17. If cos²θ / (cot²θ - cos²θ) = 3, 0° < θ < 90°, then the value of cot θ + cosec θ is :
    यदि cos²θ / (cot²θ - cos²θ) = 3, 0° < θ < 90° है, तो cot θ + cosec θ का मान क्या होगा?
    (A) (a) √3

    (B) (b) √3/2

    (C) (c) 2√3

    (D) (d) 3√3/4
    ✅ Answer & Explanation

    Sahi jawab: A) (a) √3

    Explanation: Method 1 (Angle Evaluation Method): Step 1: Cross multiply karke angle determine karte hain: $$\cos^2\theta = 3(\cot^2\theta - \cos^2\theta)$$ $$\cos^2\theta = 3\cos^2\theta(\csc^2\theta - 1) = 3\cos^2\theta \cdot \cot^2\theta$$ $$\cot^2\theta = \frac{1}{3} \implies \cot\theta = \frac{1}{\sqrt{3}} \implies \theta = 60^\circ$$ Step 2: $\cot\theta + \csc\theta$ me $\theta = 60^\circ$ substitute karein: $$\cot 60^\circ + \csc 60^\circ = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$$ Correct option (a) √3 hai. Method 2 (Direct Identity Method): $$\cot\theta + \csc\theta = \frac{\cos\theta + 1}{\sin\theta}$$ $\theta = 60^\circ$ rakhne par: $$\frac{\frac{1}{2} + 1}{\frac{\sqrt{3}}{2}} = \frac{\frac{3}{2}}{\frac{\sqrt{3}}{2}} = \frac{3}{\sqrt{3}} = \sqrt{3}$$ Sahi uttar (a) hai.
18. If sin (A - B) = 1/2 and cos (A + B) = 1/2, where A > B > 0°, and A + B is an acute angle, then the value of A is :
    यदि sin (A - B) = 1/2 और cos (A + B) = 1/2 है, जहाँ A > B > 0° है, और A + B एक न्यून कोण है, तो A का मान क्या होगा?
    (A) (a) 45°

    (B) (b) 30°

    (C) (c) 15°

    (D) (d) 75°
    ✅ Answer & Explanation

    Sahi jawab: A) (a) 45°

    Explanation: Method 1 (System of Linear Equations Method): Step 1: Standard trigonometric angles se equations banate hain: - $\sin(A - B) = \frac{1}{2} = \sin 30^\circ \implies A - B = 30^\circ \quad \text{--- (समीकरण 1)}$ - $\cos(A + B) = \frac{1}{2} = \cos 60^\circ \implies A + B = 60^\circ \quad \text{--- (समीकरण 2)}$ Step 2: Dono equations ko add karke $A$ solve karein: $$(A - B) + (A + B) = 30^\circ + 60^\circ$$ $$2A = 90^\circ \implies A = 45^\circ$$ Correct option (a) 45° hai. Method 2 (Inspection Method): Sum $= 60^\circ$, Difference $= 30^\circ$. $$A = \frac{60^\circ + 30^\circ}{2} = 45^\circ, \quad B = \frac{60^\circ - 30^\circ}{2} = 15^\circ$$ Sahi uttar (a) hai.
19. If sin (A + B) = √3/2 and tan (A - B) = 1/√3, then (2A + 3B) is equal to :
    यदि sin (A + B) = √3/2 और tan (A - B) = 1/√3 है, तो (2A + 3B) किसके बराबर होगा?
    (A) (a) 120°

    (B) (b) 135°

    (C) (c) 130°

    (D) (d) 125°
    ✅ Answer & Explanation

    Sahi jawab: B) (b) 135°

    Explanation: Method 1 (Angle Pair Solving Method): Step 1: Diye gaye ratios se angles nikalte hain: - $\sin(A + B) = \frac{\sqrt{3}}{2} = \sin 60^\circ \implies A + B = 60^\circ$ - $\tan(A - B) = \frac{1}{\sqrt{3}} = \tan 30^\circ \implies A - B = 30^\circ$ Step 2: $A$ aur $B$ solve karte hain: $$2A = 60^\circ + 30^\circ = 90^\circ \implies A = 45^\circ$$ $$B = 60^\circ - 45^\circ = 15^\circ$$ Step 3: $(2A + 3B)$ evaluate karte hain: $$2A + 3B = 2(45^\circ) + 3(15^\circ) = 90^\circ + 45^\circ = 135^\circ$$ Correct option (b) 135° hai. Method 2 (Direct Split Method): $$2A + 3B = 2(A + B) + B = 2(60^\circ) + 15^\circ = 120^\circ + 15^\circ = 135^\circ$$ Sahi uttar (b) hai.
20. For α and β both being acute angles, it is given that sin (α + β) = 1, cos(α - β) = 1/2. The values of α and β are :
    α और β दोनों न्यून कोण हैं, दिया गया है कि sin (α + β) = 1, cos(α - β) = 1/2 है। α और β के मान हैं :
    (A) (a) 75°, 15°

    (B) (b) 45°, 15°

    (C) (c) 75°, 45°

    (D) (d) 60°, 30°
    ✅ Answer & Explanation

    Sahi jawab: A) (a) 75°, 15°

    Explanation: Method 1 (Standard Angles System Method): Step 1: Equations frame karte hain: - $\sin(\alpha + \beta) = 1 = \sin 90^\circ \implies \alpha + \beta = 90^\circ$ - $\cos(\alpha - \beta) = \frac{1}{2} = \cos 60^\circ \implies \alpha - \beta = 60^\circ$ Step 2: $\alpha$ aur $\beta$ solve karte hain: $$2\alpha = 90^\circ + 60^\circ = 150^\circ \implies \alpha = 75^\circ$$ $$\beta = 90^\circ - 75^\circ = 15^\circ$$ Correct option (a) 75°, 15° hai. Method 2 (Option Verification Method): Options check karein: (a) $75^\circ + 15^\circ = 90^\circ \implies \sin 90^\circ = 1$, aur $75^\circ - 15^\circ = 60^\circ \implies \cos 60^\circ = 1/2$ (Direct Match). Sahi uttar (a) hai.

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