To determine

To show:

ddxsin2⁡x1 + cot⁡x+cos2⁡x1 + tan x= -cos⁡2x

Solution:

ddxsin2⁡x1 + cot⁡x+cos2⁡x1 + tan x= -cos⁡2x

Explanation

1) Concept:

Simplify the given function by using trigonometric identities and then apply rules of differentiation.

2) Formula:

(i)

cotx=cosxsin x

(ii)

tanx=sin xcos x

(iii)

a3+b3=a+ba2-ab+b2

(iv) Addition rule:

ddx fx+gx=ddxfx+ddx[gx]

(v) Product rule:

ddxfx. gx=fx.ddxgx+gx.ddxfx

(vi)

ddxsin⁡x=cos⁡x

(vii)

ddxcosx=-sinx

(viii)

cos2⁡x-sin2⁡x=cos2x

3) Given:

ddxsin2⁡x1 + cot⁡x+cos2⁡x1 + tan x

4) Calculations:

By using identity of cot⁡x and tan⁡x in terms of sin⁡x and cos⁡x,

sin2⁡x1 + cot⁡x+cos2⁡x1 + tan x

= sin2⁡x1 + cosxsin x+cos2⁡x1 + sin xcos x

= sin3⁡xsinx+cosx +cos3⁡xcosx+sinx

= sin3⁡x+cos3⁡xsinx+cosx 

= sinx+cosxsin2⁡x+cos2⁡x-sinx cosxsinx+cosx 

= sin2⁡x+cos2⁡x-sinx cosx

=1-sin⁡xcos⁡x

Applying derivative on both sides,

ddxsin2⁡x1 + cot⁡x+cos2⁡x1 + tan x= ddx1-sin⁡xcos⁡x

=ddx 1-ddxsin⁡xcos⁡x

=0- sin⁡xddxcos⁡x+ cosx ddxsinx

= -sinx -sinx+cosx cosx

= - -sin2⁡x+cos2⁡x

= -cos2x

Conclusion:

ddxsin2⁡x1 + cot⁡x+cos2⁡x1 + tan x= -cos⁡2x