Question

if 2sin x-1=0 then show that sec x+tanx=√3​

Answer 1

Given :

2sinx - 1 = 0

To prove :

secx + tanx = √3

Proof :

We have ,

=> 2sinx - 1 = 0

=> 2sinx = 1

=> sinx = 1/2

=> sinx = sin30°

=> x = 30°

Thus ,

=> secx + tanx = sec30° + tan30°

=> secx + tanx = 2/√3 + 1/√3

=> secx + tanx = (2 + 1)/√3

=> secx + tanx = 3/√3

=> secx + tanx = √3

Hence proved .

Answer 2

Given 

2secx+tanx=1

⟹cosx2+cosxsinx=1

                 (secx=cosx1,tanx=cosxsinx)

⟹2+sinx=cosx

⟹cosx−sinx=2

⟹21cosx−21sinx=1

⟹cos4πcosx−sin4πsinx=cos0

⟹cos(4π+x)=cos0

                         (cos(A+B)=cosAcosB−sinAsinB)

⟹cos(4π+x)=cos0

⟹4π+x=2nπ

      

                         (cosk=0⟹k=2nπ)

⟹x=2nπ−4π