Question

# If $$x = n \pi- tan^{-1} 3$$ is a solution of the equation $$12 tan 2x + \displaystyle \frac{\sqrt{10}}{cos x} + 1 = 0$$ then

A
n is any integer
B
n is an even integer
C
n is a positive integer
D
n is an odd integer

Solution

## The correct option is C n is an odd integerLet n is evenThen for $$x=n\pi -tan^{ -1 }3$$$$12tan2x+\cfrac { \sqrt { 10 } }{ cosx } +1$$ $$=12tan\left( 2n\pi -2tan^{ -1 }3 \right) +\sqrt { 10 } sec\left( n\pi -tan^{ -1 }3 \right) +1$$$$=-12tan\left( 2tan^{ -1 }3 \right) +\sqrt { 10 } sec\left( tan^{ -1 }3 \right) +1$$ $$=-12tan\left( \pi +tan^{ -1 }\cfrac { 2\times 3 }{ 1-{ 3 }^{ 2 } } \right) +\sqrt { 10 } sec\left( sec^{ -1 }\left( \sqrt { 1+{ 3 }^{ 2 } } \right) \right) +1$$ $$=-12tan\left( \pi +tan^{ -1 }\left( -\cfrac { 3 }{ 4 } \right) \right) +\sqrt { 10 } sec\left( sec^{ -1 }\left( \sqrt { 10 } \right) \right) +1$$ $$=-12tantan^{ -1 }\left( -\cfrac { 3 }{ 4 } \right) +10+1=12\times \cfrac { 3 }{ 4 } +10+1=20$$And for odd n$$12tan2x+\cfrac { \sqrt { 10 } }{ cosx } +1$$$$=12tan\left( 2n\pi -2tan^{ -1 }3 \right) +\sqrt { 10 } sec\left( n\pi -tan^{ -1 }3 \right) +1$$$$=-12tantan^{ -1 }\left( -\cfrac { 3 }{ 4 } \right) -\sqrt { 10 } sec\left( sec^{ -1 }\left( \sqrt { 10 } \right) \right) +1\\ =9-10+1=0$$Hence n is oddMathematics

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