Question

How to prove?

`(1+("cosec" \ A*tanbeta)^2)/(1+("cosec" \ C*tanbeta)^2)=(1+(cotA*sinbeta)^2)/(1+(cotC*sinbeta)^2)`

Answer 1

`LHS=(1+(cscA*tanbeta)^2)/(1+(cscC*tanbeta)^2)`

`=(1+csc^2Atan^2beta)/(1+csc^2Ctan^2beta)`

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

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

`=(cos^2beta+sin^2beta+cot^2Asin^2beta)/(cos^2beta+sin^2beta+cot^2C*sin^2beta)`

`=(1+(cotAsinbeta)^2)/(1+(cotCsinbeta)^2)=RHS`

Answer 2

Please see below.

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 :

`LHS=(1+(cscAtanB)^2)/(1+(cscCtanB)^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)`

`color(white)(LHS)=(color(blue)(1)+cot^2Asin^2B)/(color(blue)(1)+cot^2Csin^2B)`

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

`LHS=RHS`