Question

If \[\sec \theta +\tan \theta =m(>1)\] , then find the value of \[\sin \theta \] is \[\left( {{0}^{\circ }}<\theta <{{90}^{\circ }} \right)\].

Answer 1

Hint: In this type of question, we can use the trigonometric relations to convert the cosec, sec, tan and cot functions in terms of sin and cos functions which can be done using the following relations

Complete step-by-step answer:

\[\begin{align}

  & \tan x=\dfrac{\sin x}{\cos x} \\

 & \cot x=\dfrac{\cos x}{\sin x} \\

 & \cos ecx=\dfrac{1}{\sin x} \\

 & \sec x=\dfrac{1}{\cos x} \\

\end{align}\]

Also, the trigonometric equation that is used in the solution is as follows

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]



As mentioned in the question, we have to prove the given statement.

Now, as mentioned in the hint, we have to first convert every involved trigonometric function into cos and sin functions and then we can proceed further.

As mentioned in the question, we have to find the value of \[\sin \theta \] .

Now, as we have to find the value of \[\sin \theta \] , hence, we should convert the entire equation in the terms of sin function which can be done using the trigonometric relations that are mentioned in the hint, as follows

\[\begin{align}

  & \dfrac{1}{\cos \theta }+\dfrac{\sin \theta }{\cos \theta }=m \\

 & \dfrac{1+\sin \theta }{\cos \theta }=m \\

 & 1+\sin \theta =m\cos \theta \\

\end{align}\]

Now, on squaring both the side of the equation, we get the following result

\[{{\left( 1+\sin \theta \right)}^{2}}={{m}^{2}}{{\cos }^{2}}\theta \]

Now, we can use the trigonometric equation which is given in the hint to get the solution as follows

\[\begin{align}

  & {{\left( 1 \right)}^{2}}+{{\sin }^{2}}\theta +2\sin \theta ={{m}^{2}}{{\cos }^{2}}\theta \\

 & 1+{{\sin }^{2}}\theta +2\sin \theta ={{m}^{2}}\left( 1-{{\sin }^{2}}\theta \right) \\

 & 1+{{\sin }^{2}}\theta +2\sin \theta ={{m}^{2}}+{{m}^{2}}\left( {{\sin }^{2}}\theta \right) \\

 & \left( 1-{{m}^{2}} \right){{\sin }^{2}}\theta +2\sin \theta +\left( 1-{{m}^{2}} \right)=0 \\

\end{align}\]

Now, on solving the given quadratic equation in sin function, we get the following result

\[\begin{align}

  & \left( 1-{{m}^{2}} \right){{\sin }^{2}}\theta +2\sin \theta +\left( 1-{{m}^{2}} \right)=0 \\

 & \sin \theta =\dfrac{-2\pm \sqrt{{{2}^{2}}-4\left( 1-{{m}^{2}} \right)\left( 1-{{m}^{2}}

\right)}}{2\left( 1-{{m}^{2}} \right)} \\

 & \sin \theta =\dfrac{-2\pm \sqrt{4-4\left( 1-{{m}^{2}} \right)\left( 1-{{m}^{2}}

\right)}}{2\left( 1-{{m}^{2}} \right)} \\

 & \sin \theta =\dfrac{-1\pm \sqrt{1-\left( 1-{{m}^{2}} \right)\left( 1-{{m}^{2}} \right)}}{\left(

1-{{m}^{2}} \right)} \\

 & \sin \theta =\dfrac{-1\pm \sqrt{1-{{\left( 1-{{m}^{2}} \right)}^{2}}}}{\left( 1-{{m}^{2}}
\right)} \\

 & \sin \theta =\dfrac{-1\pm \sqrt{1-\left( 1+{{m}^{4}}-2{{m}^{2}} \right)}}{\left( 1-{{m}^{2}}

\right)} \\

 & \sin \theta =\dfrac{-1\pm \sqrt{\left( 2-{{m}^{2}} \right){{m}^{2}}}}{\left( 1-{{m}^{2}}

\right)} \\

 & \sin \theta =\dfrac{-1\pm m\sqrt{\left( 2-{{m}^{2}} \right)}}{\left( 1-{{m}^{2}} \right)} \\

\end{align}\]

Now, as it is given as \[\left( {{0}^{\circ }}<\theta <{{90}^{\circ }} \right)\] , hence, the value of \[\sin \theta \] is \[\sin \theta =\dfrac{m\sqrt{\left( 2-{{m}^{2}} \right)}-1}{\left( 1-{{m}^{2}} \right)}\] .

Note: The students can make an error if they don’t know the relations that are mentioned in the hint which are as follows

\[\begin{align}

  & \tan x=\dfrac{\sin x}{\cos x} \\

 & \cot x=\dfrac{\cos x}{\sin x} \\

 & \cos ecx=\dfrac{1}{\sin x} \\

 & \sec x=\dfrac{1}{\cos x} \\

\end{align}\]

The students should also know the following relation as without it the solution would be difficult to get. The relation is as follows

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]