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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)

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)

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Grade:11

2 Answers

bloodthirstydracula
15 Points
4 years ago
Answer: 2
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
3 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

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