Question

tan 3A • tan 7A = 1 so A =?​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

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

Given that,

$\rm \: tan3A. \: tan7A = 1$

can be rewritten as

$\rm \: tan3A = \dfrac{1}{tan7A}$

$\rm \: tan3A = cot7A$

can be rewritten as

$\rm \: tan3A =tan (90 \degree \: - \: 7A)$

$\rm \: 3A = 90 \degree \: - \: 7A$

$\rm \: 3A + 7A= 90 \degree \:$

$\rm \: 10A= 90 \degree \:$

$\rm\implies \:\boxed{ \rm{ \:\rm \: A= 9 \degree \: \: }}$

$\rule{190pt}{2pt}$

Alternative Method for general solution :-

$\rm \: tan3A. \: tan7A = 1$

$\rm \: \dfrac{sin3A}{cos3A} \times \dfrac{sin7A}{cos7A} = 1$

$\rm \: \dfrac{sin3A \: sin7A}{cos3A \: cos7A} = 1$

$\rm \: cos7A \: cos3A = sin7A \: sin3A$

$\rm \: cos7A \: cos3A - sin7A \: sin3A = 0$

$\rm \: cos(7A + 3A) = 0$

$\rm \: cos10A = 0$

$\rm\implies \:10A = (2n + 1)\dfrac{\pi}{2} \: \: \forall \: n \in \: Z$

$\rm\implies \:\boxed{ \rm{ \:A = (2n + 1)\dfrac{\pi}{20} \: \: \forall \: n \in \: Z \: }} \:$

$\rule{190pt}{2pt}$

Additional Information :-

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

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$