Trigonometry Important Questions for SSC CGL 2017 Exam

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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