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Trigonometry Important Questions for SSC CGL 2017 Exam

Trigonometry Important Questions for SSC CGL 2017 Exam

Trigonometry Important Questions for SSC CGL 2017 Exam

1.If  tan θ + cot θ = 2, then value of θ is

  1. 45°
  2. 60°
  3. 90°
  4. 30°

Answer : 1. 45°

Solution:

tan θ + cotθ = 2

⇒ tan θ  + \frac {1}{tan \theta } = 2

\frac { tan^2 \theta + 1 }{ tan \theta } = 2

⇒ tan2 θ + 1= 2 tan θ

⇒ (tan2 θ - 2 tan θ + 1) = 0

⇒ (tan θ - 1)2 = 0

⇒ tan θ - 1 = 0

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°

 

2. If Cos πx = x2 - x+\frac{5}{4} , then the value of x will be

  1. 0
  2. 1
  3. -1
  4. 2

Answer : 2. 1

solution :

⇒ Cos πx = x2 - x + \frac{5}{4}

⇒ Cos πx = x2 - 2.x.\frac{1}{2} + \frac{1}{4} - \frac{1}{4} + \frac{5}{4}

⇒ Cos πx = (x - \frac{1}{4})2 + 1 > 1

 

3. The numerical value of \frac{1}{ cot^2 63^\circ } - Sec2 27° + \frac{1}{Sin^2 63^\circ} - Cosec2 27° is

  1. 1
  2. 2
  3. -1
  4. 0

Answer : 2. 0

Solution :

\frac{1}{cot^2 63^\circ } - Sec 2 27° + \frac{1}{Sin^2 63^\circ} - Cosec2 27°

= 1 + tan2 63° - Sec2 27° +Cos2 63° - Cosec2 27°

= 1 + tan2 (90°-27°) - sec2 27° + Cosec2 (90°-27°) - Cosec2 27°

= 1 + Cot2 27° - Sec2 27° + Sec2 27° - Cosec2 27°

= Cosec2 27° - Cosec2 27°

= 0

 

4. If x = \frac{ Cos \theta }{ 1 - Sin\theta }, then \frac{ Cos\theta }{ 1 + Sin\theta } is equal to

  1. x - 1
  2. \frac{1}{ x }
  3. \frac{1}{ x + 1}
  4. \frac{1}{cot^2}

Answer ; 2. \frac{1}{ x }

Solution:

⇒ x = \frac{cos\theta }{ 1-sin\theta }

⇒ x = \frac{cos\theta }{ 1-sin\theta } x \frac{ 1+Sin\theta }{ 1+Sin\theta }

⇒ x = Cos θ \frac{1+Sin\theta }{ 1^2 - Sin^2\theta }

⇒ x = Cos θ \frac{1+Sin\theta }{ Cos^2\theta }     [  1-Sin2 θ = Cos2 θ ]

⇒ x = \frac{ 1+Sin \theta }{ Cos\theta }

\frac{ Cos\theta }{ 1+Sin\theta } = \frac{1}{ x }

 

5. If Sin\frac{\pi }{ x } = x2 - 2x = 2, then the value of x is

  1. 0
  2. 1
  3. -1
  4. None of these

Answer : 2. 1

Solution ;

Putting x = 1

⇒ Sin\frac{\pi }{ x } = 1 - 2+2

⇒ Sin\frac{\pi }{ x } = 1

 

6. The value of \frac{Sin 43^\circ}{ Cos47^\circ } + \frac{ Cos 19^\circ }{ Sin 71^\circ } - 8 Cos2 60° is

  1. 0
  2. 1
  3. 2
  4. -1

Answer :1. 0

Solution :

\frac{ Sin 43^\circ }{ Cos 47^\circ } + \frac{ Cos 19^\circ }{ Sin 71^\circ } - 8 Cos2 60° = \frac{SIn 43^\circ}{ Cos( 90^\circ-43^\circ } + \frac{ Cos 19^\circ }{ Sin (90^\circ - 19^\circ) } - 8 Cos2 60°

\frac{Sin 43^\circ}{ Sin 43^\circ } + \frac{ Cos 19^\circ }{ Sin 19^\circ } - 8 x \frac{ 1 }{ 4 }

⇒ 1 + 1 - 2 = 0

 

7. The  value of Cos2 30° + Sin2 60° + Tan2 45° + Sec2 60° + Cos 0° is

  1. 4\frac{ 1 }{ 2 }
  2. 5\frac{ 1 }{ 2 }
  3. 6\frac{ 1 }{ 2 }
  4. 7\frac{ 1 }{ 2 }

Answer : 4. 7\frac{ 1 }{ 2 }

Solution : Cos2 30°+Sin2 60°+Tan2 45° + Sec2 60° + Cos 0°

= (\frac { \sqrt { 3 } }{ 2 })2+ (\frac{ \sqrt {3} }{ 2 })2 + 1+ 22 + 1

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

= \frac{ 3+3+4+16+4 }{ 4 }

= \frac{ 30 }{ 4 }

= \frac{ 15 }{ 2 }

 

7. If Cos x + Cos2 x = 1, then Sin2 x + 2 Sin6 x + Sin4 x is equals to

  1. 0
  2. 3
  3. 2
  4. 1

Answer : 4. 1

solution :

⇒ Cos x + Cos2 x = 1

⇒ Cos x = 1-Cos 2 x

⇒ Cos x = Sin 2x

= Sin8 x + 2 Sin6 x + Sin  x

= (Sin 4 x + Sin2 x)2

= (Cos2 x + Sin2 x)

= 1

8. if x = p Cosec θ and y = q Cot θ then value of \frac{ x^2 }{ p^2 }- \frac{ y^2 }{ q^2 } is

  1. Sin2 θ
  2. 2 Tan θ
  3. 1
  4. 0

Answer : 3. 1

Solution :

\frac{ x^2 }{ q^2 } = Cot2 θ

Now,

\frac{ x^2 }{ p^2 }- \frac{ y^2 }{ q^2 } = Cosec2 θ - Cot2 θ

\frac{ x^2 }{ p^2 }- \frac{ y^2 }{ q^2 } = 1

9. The value of Sin 22°+Sin2 68°+Cot2 30° is

  1. 4
  2. 3
  3. \frac{ 3 }{ 4 }
  4. \frac{ 5 }{ 4 }

Answer : 1. 4

Solution :

Sin2 22°+Sin2 68°+ Cot2 30°

=  Sin (90°-68°)+Sin2 68°+ Cot2 30°

=  Cos2 68°+Sin2 68°+ Cot2 30°

= 1+Cot2 30°

= 1+3 = 4

10.  If \frac{ Sec \theta + Tan\theta }{ Sec\theta - Tan\theta } = 2\frac{ 51 }{ 79 }, then value of Sin θ is

  1. \frac{ 39 }{ 74 }
  2. \frac{ 65 }{ 144 }
  3. \frac{ 35 }{ 72 }
  4. \frac{ 91 }{ 144 }

Answer : 2. \frac{ 65 }{ 144 }

solution :

By Using Componendo and Dividendo,

\frac{ Sec\theta+ Tan\theta }{ Sec\theta - Tan\theta} = \frac{ 209 }{ 79 }

\frac{ (Sec\theta+ Tan\theta)+(Sec\theta - Tan\theta) }{ (Sec\theta + Tan\theta) - (Sec\theta -Tan\theta) } = 2\frac{ 51 }{ 79 }

\frac{ 2 Sec\theta }{ 2 Tan\theta }  = \frac{ 288 }{ 130 }

\frac{ Sec\theta }{ Tan\theta }  = \frac{ 144 }{ 65 }

⇒ Sin θ = \frac{ 65 }{ 144 }

 

11. The value of the followng :

Cos 24°+Cos 55°+Cos 125°+Cos 204°+Cos 300°

is

  1. \frac { \sqrt { 3 } }{ 2 }
  2. 2
  3. 1
  4. \frac{ 1 }{ 2 }

Answer :   4. \frac{ 1 }{ 2 }

Solution :

Cos 24°+Cos 55°+Cos 125°+Cos 204°+Cos 300°

= Cos 24°+Cos 55°+Cos (180°-55°)+Cos (180°-24°)+Cos (360°-60°) [ Cos (180°-θ) = -Cosθ, Cos (360°-θ) = Cosθ]

= Cos 24°+ Cos 55°- Cos 55°- Cos 24°+Cos 60°

= Cos 60° =  \frac{ 1 }{ 2 }

12. If Tan A + Cot A = 2, then the value of Tan10 A + Cot10 A is

  1. 4
  2. 2
  3. 6
  4. 1

Answer : 2. 2

Solution : Tan A + Cot A = 2

⇒ TanA  \frac{ 1 }{ Tan A } = 2

\frac{ Tan^2 A+1 }{ Tan A } = 2

⇒ Tan2 2A= 1 - 2Tan A = 0

⇒ (TanA - 1) 2 = 0

⇒ Tan A = 1 or Cot A =1

Thus Tan10 A + Cot10 A= 2

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