Question

(1-tanx)^2+(1-cotx)^2=(secx-cosecx)^2

Answer 1

(1-tanx)^2+(1- cotx)^2

=(1-2tanx+tan^2x)+(1-2cotx+cot^2x)

=sec^2x+csc^2x-2(tanx+cotx)

=sec^2x+csc^2x-2(sinx/cosx+cosx/sinx)

=sec^2x+csc^2x-2[(sin^2x+cos^2x)/(cosx*sinx)]

=sec^2x+csc^2x-2[1/(cosx*sinx)]

=sec^2x+csc^2x-2[(1/cosx)*(1/sinx)]

=sec^2x+csc^2x-2secx*cscx

=(secx-cscx)^2

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Answer 2

The trigonometric identities are:

$\sin^2 x+\cos^2x=1$

$1+\tan^2 x=\sec^2x$

$1+\cot^2 x=\csc^2x$

Explanation:

To prove : $(1-\tan x)^2+(1-\cot x)^2=(\sec x-\csc x)^2$

Consider LHS : $(1-\tan x)^2+(1-\cot x)^2$

Using identity : $(a-b)^2=a^2+b^2-2ab$, we get

$1^2+\tan^2 x-2\tan x+1^2+\cot^2 x-2\cot x$

$=(1+\tan^2 x)-2\tan x+(1+\cot^2 x)-2\cot x$

$=\sec^2x-2\tan x+\csc^2x -2\cot x$

[$\because\ 1+\tan^2 x=\sec^2x\ \&\ 1+\cot^2 x=\csc^2x$]

$=\sec^2x+\csc^2x-2(\tan x -\cot x)$

$=\sec^2x+\csc^2x-2(\dfrac{\sin x}{\cos x} -\dfrac{\cos x}{\sin x})$

$=\sec^2x+\csc^2x-2(\dfrac{\sin^2 x+\cos^2x}{\sin x\cos x})$

$=\sec^2x+\csc^2x-2(\dfrac{1}{\sin x\cos x})$

[$\because\ \sin^2 x+\cos^2x=1$]

$=\sec^2x+\csc^2x-2(\csc x\sec x)$

$=(\sec x-\csc x)^2$ → RHS

Hence, Proved .

# Learn more :

(1+cotx - cosecx)(1+tanx+secx)= 2

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