Question

Solve :- tanmx + cotpx = 0

Please explain with full explanation. ​

Answer 1

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

Given Trigonometric equation is

$\rm :\longmapsto\:tanmx + cotpx = 0$

can be rewritten as

$\rm :\longmapsto\:\dfrac{sin \: mx}{cos \: mx} + \dfrac{cos \: px}{sin \: px} = 0$

$\rm :\longmapsto\:\dfrac{sin \: mx \: sin \: px + \: cos \: px \: cos \: mx}{cos \: mx}= 0$

can be rewritten as

$\rm :\longmapsto\:sin \: mx \: sin \: px + \: cos \: px \: cos \: mx= 0$

We know,

$\boxed{ \tt{ \: cosx \: cosy \: + \: sinx \: siny \: = \: cos(x - y) \: \: }}$

So, using this identity, we get

$\rm :\longmapsto\:cos(px - mx) = 0$

$\rm :\longmapsto\:cos(p - m)x = 0$

We know,

$\boxed{ \tt{ \: cosx = 0 \bf\implies \: \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z}}$

So, using this, we get

$\bf\implies \:(p - m)x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z$

$\red{\bf\implies \:\boxed{ \tt{ \: x = (2n + 1)\dfrac{\pi}{2(p - m)}\: \forall \: n \in \: Z}}}$

More to Know :-

$\purple{\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}}$

Answer 2

Answer:

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

Given Trigonometric equation is

$\rm :\longmapsto\:tanmx + cotpx = 0$

can be rewritten as

$\rm :\longmapsto\:\dfrac{sin \: mx}{cos \: mx} + \dfrac{cos \: px}{sin \: px} = 0$

$\rm :\longmapsto\:\dfrac{sin \: mx \: sin \: px + \: cos \: px \: cos \: mx}{cos \: mx}= 0$

can be rewritten as

$\rm :\longmapsto\:sin \: mx \: sin \: px + \: cos \: px \: cos \: mx= 0$

We know,

$\boxed{ \tt{ \: cosx \: cosy \: + \: sinx \: siny \: = \: cos(x - y) \: \: }}$

So, using this identity, we get

$\rm :\longmapsto\:cos(px - mx) = 0$

$\rm :\longmapsto\:cos(p - m)x = 0$

We know,

$\boxed{ \tt{ \: cosx = 0 \bf\implies \: \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z}}$

So, using this, we get

$\bf\implies \:(p - m)x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z$

$\red{\bf\implies \:\boxed{ \tt{ \: x = (2n + 1)\dfrac{\pi}{2(p - m)}\: \forall \: n \in \: Z}}}$

More to Know :-

$\purple{\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}}$