# How to prove?

## $\left(1 + {\left(\text{cosec" \ A*tanbeta)^2)/(1+("cosec} \setminus C \cdot \tan \beta\right)}^{2}\right) = \frac{1 + {\left(\cot A \cdot \sin \beta\right)}^{2}}{1 + {\left(\cot C \cdot \sin \beta\right)}^{2}}$

Jul 14, 2018

$L H S = \frac{1 + {\left(\csc A \cdot \tan \beta\right)}^{2}}{1 + {\left(\csc C \cdot \tan \beta\right)}^{2}}$

$= \frac{1 + {\csc}^{2} A {\tan}^{2} \beta}{1 + {\csc}^{2} C {\tan}^{2} \beta}$

=(1+(1+cot^2A)(sin^2beta)/(cos^2beta))/(1+(1+cot^2C)*((sin^2beta)/(cos^2beta))

$= \frac{\frac{{\cos}^{2} \beta + \left(1 + {\cot}^{2} A\right) {\sin}^{2} \beta}{\cancel{{\cos}^{2} \beta}}}{\frac{{\cos}^{2} \beta + \left(1 + {\cot}^{2} C\right) {\sin}^{2} \beta}{\cancel{{\cos}^{2} \beta}}}$

$= \frac{{\cos}^{2} \beta + {\sin}^{2} \beta + {\cot}^{2} A {\sin}^{2} \beta}{{\cos}^{2} \beta + {\sin}^{2} \beta + {\cot}^{2} C \cdot {\sin}^{2} \beta}$

$= \frac{1 + {\left(\cot A \sin \beta\right)}^{2}}{1 + {\left(\cot C \sin \beta\right)}^{2}} = R H S$

Jul 14, 2018

#### Explanation:

We know that ,

color(red)((1)csc^2theta-cot^2theta=1=>csc^2theta=1+cot^2theta

color(blue)((2)cos^2theta+sin^2theta=1

We take LHS :

$L H S = \frac{1 + {\left(\csc A \tan B\right)}^{2}}{1 + {\left(\csc C \tan B\right)}^{2}}$

color(white)(LHS)=(1+color(red)(csc^2A)color(brown)(tan^2B))/(1+color(red)(csc^2C)color(brown)(tan^2B))tocolor(red)(Apply(1)

color(white)(LHS)=(1+(color(red)(1+cot^2A))color(brown)(sin^2B/cos^2B))/(1+ (color(red)(1+cot^2C))color(brown)(sin^2B/cos^2B))to[because color(brown)(tantheta=sintheta/costheta)]

color(white)(LHS)= (color(blue)(cos^2B+sin^2B)+cot^2Asin^2B)/(color(blue)(cos^2B+sin^2B)+cot^2Csin^2B )tocolor(blue)( Apply(2)

$\textcolor{w h i t e}{L H S} = \frac{\textcolor{b l u e}{1} + {\cot}^{2} A {\sin}^{2} B}{\textcolor{b l u e}{1} + {\cot}^{2} C {\sin}^{2} B}$

color(white)(LHS)=(1+(cotAsinB)^2)/(1+(cotCsinB)^2

$L H S = R H S$