Question

tano/1-coto+coto/1-tano =1+tano+coto

Answer 1

GIVEN :

The equation is $\frac{tano}{1-coto}+\frac{coto}{1-tano}= 1+tano+coto$

TO FIND :

The equality $\frac{tano}{1-coto}+\frac{coto}{1-tano}= 1+tano+coto$ is true and verify.

SOLUTION :

Given equation is $\frac{tano}{1-coto}+\frac{coto}{1-tano}= 1+tano+coto$

we know that the formulae are given by :

$tanx=\frac{sinx}{cosx}$ and $cotx=\frac{cosx}{sinx}$

Now by substituting in the LHS :

 $\frac{tano}{1-coto}+\frac{coto}{1-tano}$

$=\frac{\frac{sino}{coso}}{1-\frac{coso}{sino}} + \frac{\frac{coso}{sino}}{1-\frac{sino}{coso}}$

$=\frac{\frac{sino}{coso}}{\frac{sino-coso}{sino}}+ \frac{\frac{coso}{sino}}{\frac{coso-sino}{coso}}$

[tex]=\frac{sin^2o}{coso(sino-coso)} + \frac{cos^2o}{sino(coso-sino)} [/tex]

[tex]=\frac{sin^2o}{coso(sino-coso)} - \frac{cos^2o}{sino(sino-coso)} [/tex]

$=\frac{sino}{coso}\times \frac{sino}{sino-coso} - \frac{coso}{sino}\times \frac{coso}{sino-coso}$

$=tano\times sino(\frac{1}{sino-coso}) - coto\times coso(\frac{1}{sino-coso})$

$=\frac{1}{sino-coso} [tano sino -coto coso]$

$=\frac{1}{sino-coso}[\frac{sin^3o-cos^3o}{(sino coso)}]$

By using the algebraic identity :

$a^3-b^3=(a-b)(a^2+ab+b^2)$

$=\frac{1}{sino-coso}\times \frac{(sino-coso)(sin^2o+sino coso +cos^2o)}{(sino coso)}$

$= \frac{(sin^2o+sino coso+cos^2o)}{sino coso}$

$= \frac{sino}{coso} + 1 + \frac{coso}{sino}$

$= tano + 1 +coto$

$=1+tano+coto$

= RHS

LHS = RHS

∴ $\frac{tano}{1-coto}+\frac{coto}{1-tano}= 1+tano+coto$ is true and verified.