Question

Given that for $a,b,c,d\in R,$ if $a\sec \left({200}^o\right)-c\tan \left({200}^o\right)=d$ and $b\sec \left({200}^o\right)+d\tan \left({200}^o\right)=c$, then find the value of $\left(\frac{a^2+b^2+c^2+d^2}{bd-ac}\right)\sin {20}^o$ :

Answer 1

Given,

$a\sec \left({200}^{\circ }\right)-c\tan \left({200}^{\circ }\right)=d$ and $b\sec \left({200}^{\circ }\right)+d\tan \left({200}^{\circ }\right)=c$

Consider,

$a\sec \left({200}^{\circ }\right)-c\tan \left({200}^{\circ }\right)=d$

$\frac{a}{\cos \left({200}^{\circ }\right)}-\frac{c\sin \left({200}^{\circ }\right)}{\cos \left({200}^{\circ }\right)}=d$

$\frac{a}{\cos (180+20{)}^{\circ }}-\frac{c\sin (180+20{)}^{\circ }}{\cos (180+20{)}^{\circ }}=d$

$\frac{-a}{\cos \left({20}^{\circ }\right)}-\frac{c\sin \left({20}^{\circ }\right)}{\cos \left({20}^{\circ }\right)}=d$ ....... $\left(1\right)$

Consider,

$b\sec \left({200}^{\circ }\right)+d\tan \left({200}^{\circ }\right)=c$

$\frac{b}{\cos \left({200}^{\circ }\right)}+\frac{d\sin \left({200}^{\circ }\right)}{\cos \left({200}^{\circ }\right)}=c$

$\frac{b}{\cos (180+20{)}^{\circ }}+\frac{d\sin (180+20{)}^{\circ }}{\cos (180+20{)}^{\circ }}=c$

$\frac{-b}{\cos \left({20}^{\circ }\right)}+\frac{d\sin \left({20}^{\circ }\right)}{\cos \left({20}^{\circ }\right)}=c$ ....... $\left(2\right)$

Let ${20}^{\circ }=\alpha$

Therefore (1) becomes

$\frac{-a}{\cos \alpha }-\frac{c\sin \alpha }{\cos \alpha }=d$

Squaring on both sides

$\frac{a^2}{{\cos }^2\alpha }+\frac{c^2{\sin }^2\alpha }{{\cos }^2\alpha }+2\frac{a}{\cos \alpha }\frac{c\sin \alpha }{\cos \alpha }=d^2$

$a^2+c^2{\sin }^2\alpha +2ac\sin \alpha =d^2{\cos }^2\alpha$ ...... $\left(3\right)$

Similarly, consider (2).

On squaring both sides and simplifying.

$b^2+d^2{\sin }^2\alpha -2bd\sin \alpha =c^2{\cos }^2\alpha$ ....... $\left(4\right)$

Subtracting these two i.e. $\left(3\right)-\left(4\right),$ we get

$a^2+c^2{\sin }^2\alpha +2ac\sin \alpha -b^2-d^2{\sin }^2\alpha +2bd\sin \alpha =d^2{\cos }^2\alpha -c^2{\cos }^2\alpha$

$a^2+c^2({\sin }^2\alpha +{\cos }^2\alpha )+2ac\sin \alpha -b^2-d^2({\sin }^2\alpha +{\cos }^2\alpha )+2bd\sin \alpha =0$

$a^2+c^2-b^2-d^2+2\sin \alpha (ac+bd)=0$

$(a^2+c^2)-(b^2+d^2)=-2\sin \alpha (ac+bd)$

$\frac{(a^2+c^2)-(b^2+d^2)}{-2(ac+bd)}=\sin \alpha$ ........ $\left(5\right)$

To find the value of,

$\left(\frac{a^2+b^2+c^2+d^2}{bd-ac}\right)\sin {20}^{\circ }$

Let ${20}^{\circ }=\alpha$

$\left(\frac{a^2+b^2+c^2+d^2}{bd-ac}\right)\sin \alpha$

Substitute for $\sin \alpha$ from $\left(5\right)$

$=\left(\frac{a^2+b^2+c^2+d^2}{bd-ac}\right)\left(\frac{(a^2+c^2)-(b^2+d^2)}{-2(ac+bd)}\right)$

$=\frac{(b^2+d^2{)}^2-(a^2+c^2{)}^2}{2(b^2{d}^2-a^2{c}^2)}$

$=\frac{(b^4+d^4+2b^2{d}^2)-(a^4+c^4+2a^2{c}^2)}{2(b^2{d}^2-a^2{c}^2)}$

$=\left(\frac{(b^4+d^4-a^4-c^4)+(2b^2{d}^2-2a^2{c}^2)}{2(b^2{d}^2-a^2{c}^2)}\right)$

$=\frac{(b^4+d^4-a^4-c^4)}{2(b^2{d}^2-a^2{c}^2)}+\frac{2b^2{d}^2-2a^2{c}^2}{2b^2{d}^2-2a^2{c}^2}$

$=\frac{(b^4+d^4)-(a^4+c^4)}{2(b^2{d}^2-a^2{c}^2)})+1$