Question

The relation between angle A of the trigonometric ratio is that if $\tan A=2$ then find the value of $\sec A\sin A+{{\tan }^{2}}A-\text{cosecA}.\cos A=\dfrac{11}{m}$. Find the value of “m”?

Answer 1

Hint: We have given $\tan A=2$ and we know that $\tan A=\dfrac{P}{B}$ where P is the perpendicular with respect to angle A and B is the base with respect to angle A. The equation that we have given above contains different trigonometric ratios. So, we need the hypotenuse with respect to angle A which we can calculate by Pythagoras theorem. Now, find the different trigonometric ratios given in the above problem and then substitute these values of trigonometric ratios in the given equation and find the value of m.

Complete step-by-step solution:
We have given the trigonometric ratio for angle A as follows:
$\tan A=2$
The below figure is showing the right triangle ABC right angled at B.

We know that,
$\tan A=\dfrac{P}{B}$
In the above equation, P stands for perpendicular with respect to angle A, and B stands for the base with respect to angle A.
We have given:
$\tan A=2$
Comparing this value of $\tan A$ with the above trigonometric ratio in perpendicular and base we get,
$\tan A=2=\dfrac{P}{B}$
In the above trigonometric ratio, P is 2 and B is 1 so using Pythagoras theorem we can find the hypotenuse as follows:
$\begin{align}
  & {{\left( Hypotenuse \right)}^{2}}={{\left( Perpendicular \right)}^{2}}+{{\left( Base \right)}^{2}} \\
 & \Rightarrow Hypotenuse=\sqrt{{{\left( Perpendicular \right)}^{2}}+{{\left( Base \right)}^{2}}} \\
\end{align}$
Substituting perpendicular as 2 and base as 1 in the above equation we get,
$\begin{align}
  & Hypotenuse=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\
 & \Rightarrow Hypotenuse=\sqrt{4+1}=\sqrt{5} \\
\end{align}$
We have given the equation in the above problem as follows:
$\sec A\sin A+{{\tan }^{2}}A-\text{cosecA}.\cos A=\dfrac{11}{m}$
To find the value of m, we need the value of trigonometric ratio $\sec A,\sin A,\cos ecA,\cos A$ which we are going to find as follows:
We know the value of perpendicular (P), base (B) and hypotenuse (H) with respect to angle A is 2, 1 and $\sqrt{5}$ respectively.
We know that,
$\begin{align}
  & \sin A=\dfrac{P}{H}; \\
 & \cos A=\dfrac{B}{H} \\
\end{align}$
Substituting the value of P, B and H as 2, 1 and $\sqrt{5}$ respectively in the above equation we get,
$\sin A=\dfrac{2}{\sqrt{5}}$…….Eq. (1)
$\cos A=\dfrac{1}{\sqrt{5}}$………… Eq. (2)
We know that $\sec A$ is the reciprocal of $\cos A$ so:
$\sec A=\dfrac{1}{\cos A}$
Substituting the value of $\cos A$ from the above equation we get,
$\sec A=\sqrt{5}$
We also know that $\text{cosecA}$ is the reciprocal of $\sin A$ so:
$\text{cosecA}=\dfrac{1}{\sin A}$
Substituting the value of $\sin A$ from eq. (1) we get,
$\text{cosecA}=\dfrac{\sqrt{5}}{2}$
And we have already known the value of $\tan A=2$.
Substituting these trigonometric ratios in the following equation we get,
$\sec A\sin A+{{\tan }^{2}}A-\text{cosecA}.\cos A=\dfrac{11}{m}$
Solving the left hand side of the above equation we get,
$\left( \sqrt{5} \right)\dfrac{2}{\sqrt{5}}+{{\left( 2 \right)}^{2}}-\left( \dfrac{\sqrt{5}}{2} \right).\dfrac{1}{\sqrt{5}}$
In the above equation, $\sqrt{5}$ will be cancelled from the numerator and denominator we get,
$\begin{align}
  & 2+4-\dfrac{1}{2} \\
 & =6-\dfrac{1}{2} \\
 & =\dfrac{12-1}{2}=\dfrac{11}{2} \\
\end{align}$
From the above, we have got the value of L.H.S as $\dfrac{11}{2}$ and R.H.S is given as $\dfrac{11}{m}$ so equating both of them we get,
$\dfrac{11}{2}=\dfrac{11}{m}$
The above equation holds true when $m=2$.
Hence, the value of m that we are getting is 2.

Note: The other way of solving the above problem is by simplifying the given equation first and then substitute the trigonometric ratios.
$\sec A\sin A+{{\tan }^{2}}A-\text{cosecA}.\cos A=\dfrac{11}{m}$
This is the equation given above.
Now, we know that $\sec A$ is the reciprocal of $\cos A$ and $\cos ecA$ is the reciprocal of $\sin A$ so using these relations in the above equation we get,
$\dfrac{1}{\cos A}\left( \sin A \right)+{{\tan }^{2}}A-\dfrac{1}{\sin A}.\cos A=\dfrac{11}{m}$
We also know that,
$\dfrac{\sin A}{\cos A}=\tan A$
Using the above relation in the given equation we get,
$\tan A+{{\tan }^{2}}A-\dfrac{1}{\tan A}=\dfrac{11}{m}$
Now, the value of $\tan A$ is given as 2 so substituting that value in the above equation we get,
$\begin{align}
  & 2+{{\left( 2 \right)}^{2}}-\dfrac{1}{2}=\dfrac{11}{m} \\
 & \Rightarrow 2+4-\dfrac{1}{2}=\dfrac{11}{m} \\
\end{align}$
$\Rightarrow 6-\dfrac{1}{2}=\dfrac{11}{m}$
$\begin{align}
  & \Rightarrow \dfrac{12-1}{2}=\dfrac{11}{m} \\
 & \Rightarrow \dfrac{11}{2}=\dfrac{11}{m} \\
\end{align}$
The above equation holds when $m=2$.
Hence, this alternative solution gives the same value of m as we have solved in the solution part.