Question Number 102177 by dw last updated on 07/Jul/20

Commented byRasheed.Sindhi last updated on 07/Jul/20

ac+bd=0⇒d=−((ac)/b)  c^2 +d^2 =c^2 +(−((ac)/b))^2 =1            ⇒b^2 c^2 +a^2 c^2 =b^2               c^2 (a^2 +b^2 )=b^2                   c^2 =b^2       [∵ a^2 +b^2 =1]               ⇒c=±b               ⇒d=−((a(±b))/b)=∓a  Now  ab+cd=ab+(±b)(∓a)                =ab−ab=0  ab+cd=0 ▲^⊚ _(    ⌢)

Commented bydw last updated on 07/Jul/20

Thak you sir! Good solution.

Answered by $@y@m last updated on 07/Jul/20

Let a=sin α, b=cos α           c=sin β, d=cos β  ATQ,  sin αsin  β+cos αcos β=0  ⇒cos (α−β)=0 ...(1)  Now,  ac+bd=sin αcos α+sin βcos β  =(1/2)(sin 2α+sin 2β)  =(1/2){2sin (α+β)cos (α−β)}  =0

Commented bydw last updated on 07/Jul/20

Thanks for your solution! Trigonotetric subs. is  incredible.