Prove the identity: `cos^2x*tan x = (2*sin x)/(sec x + cos x + sin^2 x*sec x)`

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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Consider RHS

2sinx/ (secx+cosx+sin2x.secx)

We can write;

Cosx   = cosx * cosx/cosx

          = cos2x / cosx

          = cos2x *secx

 

Hence RHS can be modified as;

2sinx/ (secx+cosx+sin2x.secx)   

= 2*sinx/( secx+ cos2x *secx +sin2x.secx)

Then ( secx+ cos2x *secx +sin2x.secx) = secx (1+ cos2x+ sin2x)

We know from trigonometry cos2x+ sin2x = 1

Therefore ( secx+ cos2x *secx +sin2x.secx) = secx (1+ 1)

                                                               = 2*secx

 

So;

2sinx/ (secx+cosx+sin2x.secx)    = 2*sinx/(2*secx)

                                                = sinx/secx

                                                = Sinx/(1/cosx)

                                                = sinx.cosx

                                                = sinx.cosx*cosx/cosx

                                                = cosx.cosx*(sinx/cosx)

                                                = cos2x*tanx

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The identity `cos^2x*tan x = (2*sin x)/(sec x + cos x + sin^2 x*sec x)` has to be proved.

`cos^2x*tan x`

=> `cos^2 x*(sin x/cos x)`

=> `cos x* sin x` ...(1)

`(2*sin x)/(sec x + cos x + sin^2 x*sec x)`

=> `(2*sin x)/(1/cos x + cos x + (sin^2x)/cos x) `

=> `(2*sin x*cos x)/(1 + cos^2 x + sin^2 x)`

=> `(2*sin x*cos x)/2`

=> `sin x*cos x` ...(2)

As (1) = (2) it proves that `cos^2x*tan x = (2*sin x)/(sec x + cos x + sin^2 x*sec x)`

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