Question

If the value of sec θ + tan θ is p then find the value of cosec θ.

Answer 1

Hint: Use the formulas like
$\begin{align}
  & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \text{tan}\theta =\dfrac{\sin \theta }{\cos \theta } \\
\end{align}$

Complete step-by-step answer:
Convert the given equation in sin θ and cos θ.
Then convert cos θ into sin θ using the formula
${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$

Then a quadratic equation in sin θ will form. Use the quadratic formula to find the value of sin θ. For a quadratic equation of the form
$a{{x}^{2}}+bx+c=0$

Roots are given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Then convert sin θ into cosec θ using the given below formula
$\text{cosec}\theta =\dfrac{1}{\sin \theta }$

We know that
$\begin{align}
  & \sec \theta =\dfrac{1}{\cos \theta }\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(i)} \\
 & \text{tan}\theta =\dfrac{\sin \theta }{\cos \theta }\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(ii)} \\
\end{align}$

Using the equation (i) and equation (ii) and converting the given equation into sin θ and cos θ as:
$\begin{align}
  & \,\,\,\,\,\,\,\sec \theta +\tan \theta =p \\
 & \Rightarrow \dfrac{1}{\cos \theta }+\dfrac{\sin \theta }{\cos \theta }=p \\
\end{align}$
$\Rightarrow \dfrac{1+\sin \theta }{\cos \theta }=p$
Multiplying cos θ on both side we get
$\begin{align}
  & \,\,\,\,\,\,\dfrac{1+\sin \theta }{\cos \theta }\cos \theta =p\cos \theta \\
 & \Rightarrow 1+\sin \theta =p\cos \theta \\
\end{align}$

Squaring both sides we get
$\begin{align}
  & \,\,\,\,\,{{\left( 1+\sin \theta \right)}^{2}}={{p}^{2}}{{\cos }^{2}}\theta \\
 & \Rightarrow 1+{{\sin }^{2}}\theta +2\sin \theta ={{p}^{2}}{{\cos }^{2}}\theta \,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(iii)} \\
\end{align}$

We know that
$\begin{align}
  & \,\,\,\,\,\,\,{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \\
 & \Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \\
\end{align}$
Using the above equation in equation (iii) we get
$\begin{align}
  & \,\,\,\,\,\,\,\,1+{{\sin }^{2}}\theta +2\sin \theta ={{p}^{2}}\left( 1-{{\sin }^{2}}\theta \right) \\
 & \Rightarrow \left( {{p}^{2}}+1 \right){{\sin }^{2}}\theta +2\sin \theta +1-{{p}^{2}}=0 \\
\end{align}$

Let sin θ = t. Then we have
$\left( {{p}^{2}}+1 \right){{t}^{2}}+2t+1-{{p}^{2}}=0\,\,\,\,\,\cdot \cdot \cdot \text{(iv)}$

This is a quadratic equation in t. Now solving this equation to get the value of t in terms of p using the quadratic formula. The quadratic formula says that if the quadratic equation is
$a{{x}^{2}}+bx+c=0\,\,,\,\,a\ne 0\,\,\,\,\cdot \cdot \cdot \text{(v)}$

Then its roots are found by using the formula given below
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Since equation (iv) is a quadratic equation, comparing equation (iv) with equation (v) we get
$a={{p}^{2}}+1\,\,,\,\,b=2\,\,,\,\,c=1-{{p}^{2}}$

Using the quadratic formula to get the value of t
$\begin{align}
  & t=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\left( {{p}^{2}}+1 \right)\left( 1-{{p}^{2}} \right)}}{2\left( {{p}^{2}}+1 \right)} \\
 & \Rightarrow t=\dfrac{-2\pm \sqrt{4-4\left( 1-{{p}^{4}} \right)}}{2\left( {{p}^{2}}+1 \right)} \\
 & \Rightarrow t=\dfrac{-2\pm 2\sqrt{1-\left( 1-{{p}^{4}} \right)}}{2\left( {{p}^{2}}+1 \right)} \\
 & \Rightarrow t=\dfrac{-1\pm \sqrt{{{p}^{4}}}}{{{p}^{2}}+1} \\
 & \Rightarrow t=\dfrac{-1\pm {{p}^{2}}}{{{p}^{2}}+1} \\
 & \Rightarrow t=\dfrac{-1-{{p}^{2}}}{{{p}^{2}}+1}\,\,\,\,\,\,\,\,\,\,\,\,\,\text{or }t=\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}\, \\
 & \Rightarrow t=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{or }t=1-\dfrac{2}{{{p}^{2}}+1} \\
\end{align}$
If $t=-1$ ,
$\begin{align}
  & \sin \theta =-1 \\
 & \Rightarrow \cos \theta =\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-{{(-1)}^{2}}}=\sqrt{1-1}=0 \\
\end{align}$

Then $\sec \theta $ will not be defined.
So, $t=1-\dfrac{1}{{{p}^{2}}+1}$
$\begin{align}
  & \Rightarrow \sin \theta =1-\dfrac{2}{{{p}^{2}}+1} \\
 & \Rightarrow \sin \theta =\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1} \\
\end{align}$

Using the formula
$\text{cosec}\theta =\dfrac{1}{\sin \theta }$
We get the value of $\text{cosec}\theta $as
$\begin{align}
  & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
 & \Rightarrow \text{cosec}\theta =\dfrac{{{p}^{2}}+1}{{{p}^{2}}-1} \\
\end{align}$

Hence the value of $\text{cosec}\theta $ is found to be $\dfrac{{{p}^{2}}+1}{{{p}^{2}}-1}$

Note: There is one alternate method which is quite easy and short. This method uses the formula
$\begin{align}
  & {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \\
 & \Rightarrow \left( \sec \theta -\tan \theta \right)\left( \sec \theta +\tan \theta \right)=1 \\
\end{align}$
So if $\sec \theta +\tan \theta =p$then $\sec \theta -\tan \theta =\dfrac{1}{p}$
Now subtracting second equation by first we get
$\begin{align}
  & 2\tan \theta =p-\dfrac{1}{p} \\
 & \Rightarrow \tan \theta =\dfrac{{{p}^{2}}-1}{2p} \\
 & \Rightarrow \cot \theta =\dfrac{2p}{{{p}^{2}}-1} \\
\end{align}$
We know that
 $\begin{align}
  & \text{cose}{{\text{c}}^{2}}\theta =1+{{\cot }^{2}}\theta \\
 & \Rightarrow \text{cose}{{\text{c}}^{2}}\theta =1+\dfrac{4{{p}^{2}}}{{{\left( {{p}^{2}}-1 \right)}^{2}}} \\
 & \Rightarrow \text{cose}{{\text{c}}^{2}}\theta =\dfrac{{{\left( {{p}^{2}}+1 \right)}^{2}}}{{{\left( {{p}^{2}}-1 \right)}^{2}}} \\
 & \Rightarrow \text{cosec}\theta =\dfrac{{{p}^{2}}+1}{{{p}^{2}}-1} \\
\end{align}$