If 4sinA+secA=0 then tanA= a)-2-√3 b)4+√2 c)4-√2 d)4+2√3

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Answer is (a) , divide whole equation by cosA and then u will find that it is quadratic in tan as 4tanA+1+tan^2A=0so by this apply shree dhara charya method and u will found out that the answer is A

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If 4sinA+secA=0 then tanA= a)-2-√3 b)4+√2 c)4-√2 d)4+2√3

If 4sinA+secA=0 then tanA=
a)-2-√3 b)4+√2 c)4-√2 d)4+2√3

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If 4sinA+secA=0 then tanA= a)-2-√3 b)4+√2 c)4-√2 d)4+2√3

If 4sinA+secA=0 then tanA=a)-2-√3 b)4+√2 c)4-√2 d)4+2√3

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If 4sinA+secA=0 then tanA=
a)-2-√3 b)4+√2 c)4-√2 d)4+2√3