The number of solutions of the equation \tan A+ \sec A = 2 \cos A lying in the interval [0, 2\pi ]

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    2

  • Option 4)

    3

 

Answers (1)

As we learnt in

General Solution of Trigonometric Ratios -

\sin \Theta = \sin \alpha

\Theta = n\pi + \left ( -1 \right )^{n}\alpha , n\epsilon I

- wherein

\alpha is the given angle

 

 \tan A + \sec A = 2 \cos A

\Rightarrow \frac{\sin A}{\cos A}+\frac{1}{\cos A}=2 \cos A

\Rightarrow 1 + \sin A = 2 \cos^{2} A= 2- 2 \sin^{2}A

\Rightarrow 2 \sin^{2} A + \sin A - 1<O

\Rightarrow 2 \sin^{2} A +2 \sin A - \sin A - 1=0

\Rightarrow 2 \sin A (\sin A +1) - 1 (\sin A +1 )= 0

\therefore \sin A = \frac{1}{2} \therefore A= 30^{\circ}, 150^{\circ}

 

 


Option 1)

0

Incorrect option 

Option 2)

1

Incorrect option 

Option 3)

2

correct option 

Option 4)

3

Incorrect option 

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