Question

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.
cot x sec4x = cot x + 2 tan x + tan3x

(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x

1 + sec2x sin2x = sec2x

sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x - tan2x + sec2x = 1

Answer 1

Answer:

Step-by-step explanation:

1.

cot x sec⁴ x = cot x+2 tan x +tan³x

L.H.S = cot x sec⁴x

       =cot x (sec²x)²

       =cot x (1+tan²x)²     [ ∵ sec²x=1+tan²x]

       =  cot x(1+ 2 tan²x +tan⁴x)

       =cot x+ 2 cot x tan²x+cot x tan⁴x

        =cot x +2 tan x + tan³x        [ ∵cot x tan x $=\frac{ \textrm{tan x }}{\textrm{tan x}}$ =1]

       =R.H.S

2.

(sin x)(tan x cos x - cot x cos x)=1-2 cos²x

 L.H.S =(sin x)(tan x cos x - cot x cos x)

          = sin x tan x cos x - sin x cot x cos x

           $=\textrm{sin x cos x }\times\frac{\textrm{sin x}}{\textrm{cos x} } - \textrm{sinx}\times\frac{\textrm{cos x}}{\textrm{sin x}}\times \textrm{cos x}$

           = sin²x -cos²x

           =1-cos²x-cos²x

           =1-2 cos²x

           =R.H.S

         

3.

1+ sec²x sin²x =sec²x

L.H.S =1+ sec²x sin²x

         =$1+\frac{{sin^2x}}{cos^2x}$                       [$\textrm{sec x}=\frac{1}{\textrm{cos x}}$]

         =1+tan²x                        $[\frac{\textrm{sin x}}{\textrm{cos x}} = \textrm{tan x}]$

         =sec²x

        =R.H.S

4.

$\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}} = \textrm{2 csc x}$

L.H.S=$\frac{\textrm{sinx}}{\textrm{1-cos x}} +\frac{\textrm{sinx}}{\textrm{1+cos x}}$

       $=\frac{\textrm{sinx(1+cos x)+{\textrm{sinx(1-cos x)}}}}{\textrm{(1-cos x)\textrm{(1+cos x})}}$

      $=\frac{\textrm{sinx+sin xcos x+{\textrm{sinx-sin xcos x}}}}{{(1-cos ^2x)}}$

     $=\frac{\textrm{2sin x}}{sin^2 x}$

      = 2 csc x

    = R.H.S

5.

-tan²x + sec²x=1

L.H.S=-tan²x + sec²x

        = sec²x-tan²x

        =$\frac{1}{cos^2x} -\frac{sin^2x}{cos^2x}$

        $=\frac{1- sin^2x}{cos^2x}$

        $=\frac{cos^2x}{cos^2x}$

        =1