Question

01:13:38
MCQ (100)
Q15. If 3 * tan(x - 15) = tan(x + 15). then the value of x is
(A) 30
(B)
45
(C) 60
(D) 90​

Answer 1

Answer :-

• $\sf{x =}$ $\textsf{\textbf{\red{45°}}}$

Solution :-

➪ $\sf{3tan(x-15°)=tan(x+15°)}$

➪ $\sf{\dfrac{3}{1}=\dfrac{tan(x+15°)}{tan(x-15°)}}$

By using componendo dividendo,

➪ $\sf{\dfrac{3+1}{3-1}=\dfrac{tan(x+15°)+tan(x-15°)}{tan(x+15°)-tan(x-15°)}}$

➪ $\sf{\dfrac{4}{2}=\dfrac{\dfrac{sin(x+15°)}{cos(x+15°)}+\dfrac{sin(x-15°)}{cos(x-15°)}}{\dfrac{sin(x+15°)}{cos(x+15°)}-\dfrac{sin(x-15°)}{cos(x-15°)}}}$

$\sf{➪ \:2=\dfrac{sin(x+15°).cos(x-15°)+cos(x+15°).sin(x-15°)}{sin(x+15°).cos(x-15°)-cos(x+15°).sin(x-15°)}}$

By applying,

sin(A + B) = sinA.cosB + cosA.sinB

sin(A - B) = sinA.cosB - cosA.sinB

We get,

➪ $\sf{2=\dfrac{sin(2x)}{sin(30°)}}$

➪ $\sf{sin(2x)=\cancel{2}\times \dfrac{1}{\cancel{2}}}$

We know that sin 90° = 1

➪ So, $\sf{2x = 90°}$

➪ $\sf{x =}$ $\bold{\red{45°}}$

Extra information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered}$