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Question

If cosx+sinx=12, where x(0,π), then the maximum possible value of tanx is

A
473
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B
4+73
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C
4+73
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D
473
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Solution

The correct option is C 4+73
cosx+sinx=12
Squaring both sides,
1+sin2x=14
sin2x=34

When x(0,π2)2x(0,π)
Then sin2x positive
When x(π2,π)2x(π,2π)
Then sin2x negative
So,
x(π2,π)
Now,
sin2x=342tanx1+tan2x=34
3tan2x+8tanx+3=0
tanx=8±64366tanx=4±73
As both values are negative, so both are acceptable.
Therefore the maximum possible value is,
tanx=4+73

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