Question

If a + b tan theta = sec theta and b-a tan theta = 3 sec theta , then
find the value of a^2 + b^2.​

Answer 1

Answer:

$a^2+b^2=10$

Step-by-step explanation:

easy question,

$a+btanθ$=$secθ$       .....................................(1)

$b-atanθ$=$3secθ$     ......................................(2)

squaring both (1) and (2)

we get,

$a^{2} +b^{2} tan^2θ+2abtanθ=sec^2θ$ ....................(3)

$b^{2} +a^2tan^2θ-2abtanθ=9sec^2θ$   ....................(4)

now adding both (3) & (4)

we get..

$a^2+b^2+tan^2θ(a^2+b^2)=10sec^2θ$

take $a^2+b^2$ commonly outside..

now we have..

$a^2+b^2(1+tan^2θ)=10sec^2θ$

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Identity :

$sec^2θ -tan^2θ=1$

$sec^2θ=1+tan^2θ$

_________________________

now,

$(a^2+b^2)(sec^2θ)=10sec^2θ$

the equation hence becomes,

$a^2+b^2=10$

that's the answer...