Question

Prove this ?

Answer 1

see below

Explanation:

`l tan theta+msec theta=n, l'tan theta-m'sec theta=n'`

Prove

`=>((nl'-ln')/(ml'+lm'))^2=1+((nm'+mn')/(lm'+ml'))^2`

`=>(((l tan theta+msec theta)l'-l(l'tan theta-m'sec theta))/(ml'+lm'))^2=1+(((l tan theta+msec theta)m'+m(l'tan theta-m'sec theta))/(lm'+ml'))^2`

`=>((ll' tan theta+ml'sec theta-ll'tan theta+lm'sec theta)/(ml'+lm'))^2=1+((l m'tan theta+mm'sec theta+ml'tan theta-mm'sec theta)/(lm'+ml'))^2`

`=>((cancel(ll' tan theta)+ml'sec theta-cancel(ll'tan theta)+lm'sec theta)/(ml'+lm'))^2=1+((l m'tan theta+cancel(mm'sec theta)+ml'tan theta-cancel(mm'sec theta))/(lm'+ml'))^2`

`=>((ml'sec theta+lm'sec theta)/(ml'+lm'))^2=1+((l m'tan theta+ml'tan theta)/(lm'+ml'))^2`

`=>((sec theta(ml'+lm'))/(ml'+lm'))^2=1+((tan theta(l m'+ml'))/(lm'+ml'))^2`

`=>(sec thetacancel((ml'+lm'))/cancel(ml'+lm'))^2=1+(tan thetacancel((l m'+ml'))/cancel(lm'+ml'))^2`

`=>(sec theta)^2=1+(tan theta)^2`

`=>sec ^2theta=1+tan^2 theta` , but `1+tan^2 theta=sec^2 theta`

`=>sec ^2theta=sec ^2theta`

`:. LHS=RHS`

Answer 2

Please refer to a Proof given in the Explanation Section.

Explanation:

We rewrite these eqns. as,

`l tantheta+msectheta-n=0,.................(1),` and,

`l'tantheta-m'sectheta-n'=0,........................(2).`

Using Kramer's Method to solve these eqns. for

`tantheta, and, sectheta.` we have,

`tantheta/|(m,-n),(-m',-n')|=-sectheta/|(l,-n),(l',-n')|=1/|(l,m),(l',-m')|,`

`:. tantheta={-(mn'+m'n)}/{-(lm'+l'm)}, -sectheta=(l'n-l n')/-(lm'+l'm),`

Since, `sec^2theta=1+tan^2theta,` we get,

`{(l'n-l n')/(lm'+l'm)}^2=1+{(mn'+m'n)/(lm'+l'm)}^2,` as desired!

Enjoy Maths.!