# Given that a,b,c,d ⊆R if a sec(200°) - c tan(200°)=d and b sec(200°) + dtan(200°)=c then find the value of (a^2+b^2+c^2+d^2)(sim20°)/(bd-ac)

bloodthirstydracula
15 Points
4 years ago
Solution: ​
a sec(200°) - c tan(200°)=d
a= c sin(200°) + dcos(200°)       {Multiply both sides by cos(200°) and shift csin(200°) to rhs}
-a= csin(20°) + dcos(20°)           {Subtract 180° from both cos and sin funtions (multiple angle formulae)}
Similarly
-b= ccos(20°)  + dsin(20°)
Therefore
a2 + b2 = c2 + d2                             {Square and add both sides}
-ac=dccos(20°)+c2sin(20°)    {c multipled both sides}
(-) -bd=dccos(20°)+d2sin(20°)    {d multipled both sides}
=bd-ac=(c2 + d2)sin(20°) =(a2 + b2)sin(20°)
(c2 + d2)sin(20°)/(bd-ac)=(a2 + b2)sin(20°)/(bd-ac)=1
Therefore
$\frac{(a^2+b^2+c^2+d^2)(sin20�)}{bd-ac}=2$
Vikas TU
14149 Points
4 years ago
a^2 + c^2 = b^2 + d^2
a^2 + (b secA + a tanA)^2 = b^2 + (a secA + b tanA)^2
a^2 + (b secA)^2 + (a tanA)^2 + 2(b)(secA)(a)(tanA) = b^2 + (a secA)^2 + (b tanA)^2 + 2(a)(secA)(b)(tanA)
a^2 (1 + tan^2A) + b^2 sec^2A = b^2 (1 + tan^2A) + a^2 sec^2A
Now, 1 + tan^2A = sec^2A
so LHS = RHS