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Question

If θ(π2,π2) and tan2θW, then the number of solution(s) of the equation
(1tanθ)(1+tanθ)sec2θ+2tan2θ=0 is

A
2
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B
0
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C
4
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D
Infinitely many
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Solution

The correct option is A 2
Given:
(1tanθ)(1+tanθ)sec2θ+2tan2θ=0
(1tan2θ)(1+tan2θ)+2tan2θ=0
Let tan2θ=t
(1t)(1+t)+2t=01t2+2t=02t=t21

As tan2θW
So, by observation, t=3 is the only solution.
23=321
Now,
tan2θ=3
tanθ=±3
As θ(π2,π2), so
θ=±π3

Hence, the number of solution is 2.

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