Find the value of $\frac {(a^2+b^2+c^2+d^2)(\sin 20°)}{(bd-ac)}.$

Question

Given that $a,b,c,d \in R$, if $$ a \sec(200°) - c \tan(200°) =d$$ and $$b \sec(200°) + d \tan(200°) = c$$ then find the value of $$\dfrac {(a^2+b^2+c^2+d^2)(\sin 20°)}{(bd-ac)}.$$

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My attempt:Let $\theta=200^o$ then our equations become $a \sec\theta - c \tan\theta =d$ and $b \sec\theta+ d \tan\theta= c$
Squaring and adding them,we get
$a^2\sec^2\theta+c^2\tan^2\theta-2ac\sec\theta\tan\theta+b^2\sec^2\theta+d^2\tan^2\theta+2bd\sec\theta\tan\theta=d^2+c^2$

$a^2\sec^2\theta+c^2\tan^2\theta+b^2\sec^2\theta+d^2\tan^2\theta-d^2-c^2=2ac\sec\theta\tan\theta-2bd\sec\theta\tan\theta$

I am stuck here.

Answer 1

Hint:

Method$\#1:$

Use $\sec(180^\circ+y)=-\sec y,\tan(180^\circ+y)=+\tan y$

Solve for $\sec20^\circ,\tan20^\circ$

Finally $\sin t=\dfrac{\tan t}{\sec t}\ \ \ \ (1)$

Here $t=20^\circ$

Method$\#2:$

Solve for $\sec200^\circ,\tan200^\circ$

Use $(1),$ here $t=200^\circ$

Finally $\sin(180^\circ+y)=-\sin y $