Question

asecθ + b tanθ + c = 0 and psecθ + q tanθ + r = 0 then prove that (br – qc)2 - (pc - ar)2 = ( aq - pb)2​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:a \: sec \theta + b \: tan\theta + c = 0$

and

$\rm :\longmapsto\:p\: sec \theta + q \: tan\theta + r = 0$

can be further rewritten as

$\rm :\longmapsto\:a \: sec \theta + b \: tan\theta = - \: c$

and

$\rm :\longmapsto\:p \: sec \theta + q \: tan\theta = - \: r$

Now, using Cross multiplication method, we have

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf b & \sf - c & \sf a & \sf b\\ \\ \sf q & \sf - r & \sf p & \sf q\\ \end{array}} \\ \end{gathered}$

Now, we have

$\rm :\longmapsto\:\dfrac{sec\theta}{ - br + qc} = \dfrac{tan\theta}{ - pc + ar} = \dfrac{ - 1}{aq - pb}$

On multiply by - 1, each term, we get

$\rm :\longmapsto\:\dfrac{sec\theta}{ br - qc} = \dfrac{tan\theta}{ pc - ar} = \dfrac{1}{aq - pb}$

Therefore,

$\bf\implies \:\boxed{ \tt{ \: sec\theta = \frac{br - qc}{aq - pb}}}$

and

$\bf\implies \:\boxed{ \tt{ \: tan\theta = \frac{pc - ar}{aq - pb}}}$

Now, we know that,

$\red{\rm :\longmapsto\: {sec}^{2}\theta - {tan}^{2}\theta = 1}$

On Substituting the values, evaluated above we get

$\rm :\longmapsto\: {\bigg[\dfrac{br - qc}{aq - pb} \bigg]}^{2} - {\bigg[\dfrac{pc - ar}{aq - pb} \bigg]}^{2} = 1$

$\bf\implies \: {(br - qc)}^{2} - {(pc - ar)}^{2} = {(aq - pb)}^{2}$

Hence, Proved

More to know :-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1