Question

If p and q are the roots of the equation a cosx + b sinx = c, find the value of tan(p+q)​

Answer 1

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

Given that,

p and q are the roots of the equation a cosx + b sinx = c.

Now, Consider

$\rm \: a \: sinx \: + \: b \: cosx \: = \: c$

On dividing both sides by cosx, we get

$\rm \: a \: tanx \: + \: b \: = \: c \: secx$

On squaring both sides, we get

$\rm \: (a \: tanx \: + \: b)^{2} \: = \: c ^{2} \: sec^{2} x$

$\rm \: {a}^{2} {tan}^{2}x + {b}^{2} + 2ab \: tanx \: = \: c ^{2}(1 + tan^{2} x)$

$\rm \: {a}^{2} {tan}^{2}x + {b}^{2} + 2ab \: tanx \: = \: c ^{2} + {c}^{2} tan^{2} x$

$\rm \: {a}^{2} {tan}^{2}x + {b}^{2} + 2ab \: tanx \: - \: c ^{2} - {c}^{2} tan^{2} x = 0$

$\rm \:( {a}^{2} - {c}^{2}){tan}^{2}x + 2ab \: tanx \: + \: {b}^{2} - \: c ^{2} = 0$

Since, its a quadratic in tanx, so its roots are tanp and tanq.

So,

$\rm \: tanp + tanq = \dfrac{2ab}{ {a}^{2} - {c}^{2} }$

and

$\rm \: tanp \: \times \: tanq = \dfrac{ {b}^{2} - {c}^{2} }{ {a}^{2} - {c}^{2} }$

Now, Consider

$\rm \: tan(p + q)$

$\rm \: =  \: \dfrac{tanp + tanq}{1 - tanp \times tanq}$

$\rm \: =  \: \dfrac{\dfrac{2ab}{ {a}^{2} - {c}^{2} } }{1 - \dfrac{ {b}^{2} - {c}^{2} }{ {a}^{2} - {c}^{2} } }$

$\rm \: =  \: \dfrac{\dfrac{2ab}{ {a}^{2} - {c}^{2} } }{\dfrac{ {a}^{2} - {c}^{2} - {b}^{2} + {c}^{2} }{ {a}^{2} - {c}^{2} } }$

$\rm \: =  \: \dfrac{2ab}{ {a}^{2} - {b}^{2} }$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \:tan(p + q) =  \: \dfrac{2ab}{ {a}^{2} - {b}^{2} } \: \: }}$

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Formulae Used :-

$\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: }}$

${\rm :\longmapsto\: \alpha , \beta \: are \: roots \: of \: a {x}^{2} + b {x} + c = 0, \: then}$

$\boxed{{\sf Sum\ of\ the\ roots=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\boxed{{\sf Product\ of\ the\ roots=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\boxed{\sf{  \:\rm \: tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany} \: \: }}$