Question

Prove d/dx(tanx/(1+secx))=csc²x-cotxcscx

Answer 1

First, let's find the derivative:

`f(x)=(tan\ x)/(1+sec\ x)\\\\\\f'(x)=(a'b-ab')/(b^2)\\\\\rightarrow a=tan\ x\qquad \qquad a'=\sec^2\ x\\\rightarrow b=1+sec\ x\qquad \ b'=sec\x \cdot tan\ x\\\\\\f'(x)=(sec^2\ x(1+sec\ x)-tan\ x(sec\ x\cdot tan\ x))/((1+sec\ x)^2)\\\\\\.\quad =(1)/(1+cos\ x)`

Next, manipulate the derivative to match the right hand side:

`(1)/(1+cos\ x)\bigg((1-cos\ x)/(1-cos\ x)\bigg)\\\\\\\rightarrow \quad (1-cos\ x)/(1-cos^2\ x)\\\\\\\rightarrow \quad (1-cos\ x)/(sin^2\ x)\\\\\\\rightarrow \quad (1)/(sin^2\ x)- (cos\ x)/(sin^2\ x)\\\\\\\rightarrow \quad (1)/(sin^2\ x)- (cos\ x)/(sin\ x)\cdot (1)/(sin\ x)\\\\\\\rightarrow \quad csc^2\ x-cot\ x\cdot csc\ x`

LHS = RHS
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