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Question

If 1+cot2 θ=(3+221)2, find the value of 1tan θ+sin θ1+cos θ

A
2
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B
1+2
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C
2
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D
1
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Solution

The correct option is C 2
1+cot2θ=cosec2θ=(3+221)2;
On comparing, cosecθ=3+221
3+22 can be written as
1+2+22=(1)2+(2)2+2×1×2
On comparing with (a2+b2+2ab)=(a+b)2
We can write as,
(1+2)2=1+2
So, cosec θ=1+21
cosec θ=2= cosec 45
θ=45
1tanθ+sinθ1+cosθ=1tan45+sin451+cos45
=11+121+12
=1+11+2
On simplifying we get, 2

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