Question

If \[\sec \theta +\tan \theta =p\], then find the value of \[\csc \theta \].

Answer 1

Hint: To find \[\csc \theta \], convert the whole equation into \[\sin \theta \] and square the expression. Put \[x=\sin \theta \] and solve the quadratic equation formed.

Complete step-by-step answer:

We have been given that, \[\sec \theta +\tan \theta =p\].
Let us convert \[\tan \theta \] and \[\sec \theta \] into \[\sin \theta \] and \[\cos \theta \].

We know that, \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and \[\sec \theta =\dfrac{1}{\cos \theta }\].
\[\begin{align}
  & \therefore \dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos \theta }=p \\
 & \Rightarrow \dfrac{1+\sin \theta }{\cos \theta }=p \\
\end{align}\]

We need to find \[\csc \theta \]. To find \[\csc \theta \], we need to convert everything into \[\sin \theta \]. Thus let us square both sides of the expression we get.
\[\begin{align}
  & {{\left( \dfrac{1+\sin \theta }{\cos \theta } \right)}^{2}}={{p}^{2}} \\
 & \Rightarrow \dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }={{p}^{2}} \\
\end{align}\]

We know, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
\[\therefore \dfrac{1+2\sin \theta +{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }={{p}^{2}}\]

We know, \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\]
\[\therefore {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \]

Cross multiply and simplify the expression.
\[\begin{align}
  & 1+2\sin \theta +{{\sin }^{2}}\theta ={{p}^{2}}{{\cos }^{2}}\theta \\
 & 1+2\sin \theta +{{\sin }^{2}}\theta ={{p}^{2}}\left( 1-{{\sin }^{2}}\theta \right) \\
 & 1+2\sin \theta +{{\sin }^{2}}\theta ={{p}^{2}}-{{p}^{2}}{{\sin }^{2}}\theta -(1) \\
\end{align}\]

Let us put, \[x=\sin \theta \] in equation (1).
\[\begin{align}
  & 1+2x+{{x}^{2}}={{p}^{2}}-{{p}^{2}}x \\
 & \Rightarrow 1+2x+{{x}^{2}}+p{{x}^{2}}-{{p}^{2}}=0 \\
 & {{x}^{2}}\left( {{p}^{2}}+1 \right)+2x+\left( 1-{{p}^{2}} \right)=0-(2) \\
\end{align}\]

The above equation is in the form of a quadratic equation \[a{{x}^{2}}+bx+c=0\]. Let’s compare both the equations and we get, \[a={{p}^{2}}+1\], b = 2, \[c=1-{{p}^{2}}\].
Thus substitute these values in the quadratic formula, \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
\[\begin{align}
  & x=\dfrac{-1\pm \sqrt{{{2}^{2}}-4\left( {{p}^{2}}+1 \right)\left( 1-{{p}^{2}} \right)}}{2\left( 1+{{p}^{2}} \right)} \\
 & x=\dfrac{-2\pm \sqrt{4-4\left( 1+{{p}^{2}} \right)\left( 1-{{p}^{2}} \right)}}{2\left( 1+{{p}^{2}} \right)}=\dfrac{-2\pm 2\sqrt{1-\left( 1+{{p}^{2}} \right)\left( 1-{{p}^{2}} \right)}}{2\left( 1+{{p}^{2}} \right)} \\
\end{align}\]

Cancel out 2 from numerator and denominator.
\[\begin{align}
  & x=\dfrac{-1\pm \sqrt{1-\left( {{1}^{2}}-{{\left( {{p}^{2}} \right)}^{2}} \right)}}{\left( 1+{{p}^{2}} \right)}=\dfrac{-1\pm \sqrt{1-\left( 1-{{p}^{4}} \right)}}{\left( 1+{{p}^{2}} \right)} \\
 & \because \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} \\
\end{align}\]

Similarly, \[\left( 1-{{p}^{2}} \right)\left( 1+{{p}^{2}} \right)=1-{{p}^{4}}\].
\[\therefore x=\dfrac{-1\pm \sqrt{1-1+{{p}^{4}}}}{1+{{p}^{2}}}=\dfrac{-1\pm \sqrt{{{p}^{4}}}}{1+{{p}^{2}}}=\dfrac{-1\pm {{p}^{2}}}{1+{{p}^{2}}}\]

Thus put \[x=\sin \theta \]. We get,
\[\sin \theta =\dfrac{-1\pm {{p}^{2}}}{\left( 1+{{p}^{2}} \right)}\]

Let us first consider, \[\sin \theta =\dfrac{-1+{{p}^{2}}}{\left( 1+{{p}^{2}} \right)}\].
\[\therefore \sin \theta =\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}\]

We know, \[\csc =\dfrac{1}{\sin \theta }\].
\[\therefore \csc =\dfrac{{{p}^{2}}+1}{{{p}^{2}}-1}\]

Now let us consider the case where, \[\sin \theta =\dfrac{-1-{{p}^{2}}}{1+{{p}^{2}}}\].
\[\begin{align}
  & \therefore \sin \theta =\dfrac{-\left( 1+{{p}^{2}} \right)}{\left( 1+{{p}^{2}} \right)} \\
 & \therefore \sin \theta =-1 \\
 & \csc =\dfrac{1}{\sin \theta }=-1 \\
\end{align}\]

Thus we got the value of \[\csc \theta \].

Note: Here we put, \[x=\sin \theta \], in order to make the simplification easier. If we are directly processing with \[\sin \theta \], then the solution becomes complex. Similarly, we have used basic trigonometric formulas here which you should remember.