Given that for a, b, c, d ∈ R , if a (200^ )-c (200^ )=d and b (200^ )+d (200^ )=c then (a^2+b^2+c^2+d^2)/(a c-b d)=p cosec (200^ ) find the value of p

Correct answer: 2 aligned & a (200^ ) - c (200^ ) (200^ ) =d ....(i) & b (200^ ) + d (200^ ) (200^ ) =c ....(ii) & a=d (200^ )+c (200^ ) .....(iii) & b=c (200^ )-d (200^ ) ....(iv) & a^2+b^2=c^2+d^2 & (a^2+b^2+c^2+d^2)=2(c^2+d^2) & a c=d c (200^ )+c^2 (200^ ) aligned b d=cd (200^ )-d^2 (200^ ) Also, a c-b d=(c^2+d^2) (200^ ) aligned & So (a^2+b^2+c^2+d^2)/(a c-b d)= 2(c^2+d^2) (c^2+d^2) (200^ ) =2 cosec(200^ ) & p=2 aligned

JEE Advanced

Exam: JEE Advanced
Subject: Mathematics
Chapter: Trigonometry

JEE Advanced · Mathematics · Trigonometry

Given that for a, b, c, d ∈ R , if a (200^ )-c (200^ )=d and b...

Given that for

a, b, c, d \in R

, if

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

then

\frac{a^2+b^2+c^2+d^2}{a c-b d}=p \operatorname{cosec}

\left(200^{\circ}\right)

find the value of

p \qquad

Correct answer

Step-by-step solution

\begin{aligned} & \frac{a}{\cos \left(200^{\circ}\right)}-\frac{c \sin \left(200^{\circ}\right)}{\cos \left(200^{\circ}\right)}=d ....(i)\\ & \frac{b}{\cos \left(200^{\circ}\right)}+\frac{d \sin \left(200^{\circ}\right)}{\cos \left(200^{\circ}\right)}=c ....(ii)\\ & a=d \cos \left(200^{\circ}\right)+c \sin \left(200^{\circ}\right) .....(iii)\\ & b=c \cos \left(200^{\circ}\right)-d \sin \left(200^{\circ}\right) ....(iv)\\ & \therefore a^2+b^2=c^2+d^2 \\ & \therefore\left(a^2+b^2+c^2+d^2\right)=2\left(c^2+d^2\right) \\ & \quad a c=d c \cos \left(200^{\circ}\right)+c^2 \sin \left(200^{\circ}\right) \end{aligned}

b d=\operatorname{cd} \cos \left(200^{\circ}\right)-d^2 \sin \left(200^{\circ}\right)

Also, a c-b d=\left(c^2+d^2\right) \sin \left(200^{\circ}\right)

\begin{aligned} & \text { So } \frac{a^2+b^2+c^2+d^2}{a c-b d}=\frac{2\left(c^2+d^2\right)}{\left(c^2+d^2\right) \sin \left(200^{\circ}\right)}=2 \operatorname{cosec}\left(200^{\circ}\right) \\ & \therefore p=2 \end{aligned}

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