Question

If \[\cos ec\theta +\cot \theta =k\] , then \[\cos \theta =\dfrac{{{k}^{2}}-1}{{{k}^{2}}+1}\] is?
A.True
B.False

Answer 1

Hint: -In such questions, we first have to convert the given expression into the basic form that is we should use all the relations between different trigonometric functions into sin and cos functions. The relations that will used in the conversions will be as follows
\[\begin{align}
  & \tan x=\dfrac{\sin x}{\cos x} \\
 & \cot x=\dfrac{\cos x}{\sin x} \\
 & \cos ecx=\dfrac{1}{\sin x} \\
 & \sec x=\dfrac{1}{\cos x} \\
\end{align}\]
Using the above mentioned relations, we can easily convert the given expression into the most basic form.

Complete step-by-step answer:
As mentioned in the question, we have to find whether the given expression is true or false.
Now, for this, we will first convert the given expression into the most basic form that is in the terms of sin and cos functions using the relations given in the hint as follows
\[\begin{align}
  & \cos ec\theta +\cot \theta =k \\
 & \dfrac{1}{\sin \theta }+\dfrac{\cos \theta }{\sin \theta }=k \\
 & \dfrac{1+\cos \theta }{\sin \theta }=k \\
\end{align}\]
On squaring both side of the equation, we get
\[\begin{align}
  & \dfrac{{{\left( 1+\cos \theta \right)}^{2}}}{{{\left( \sin \theta \right)}^{2}}}={{k}^{2}} \\
 & 1+{{\cos }^{2}}\theta +2\cos \theta ={{k}^{2}}(1-{{\cos }^{2}}\theta ) \\
 & \left( 1+{{k}^{2}} \right){{\cos }^{2}}\theta +2\cos \theta +1-{{k}^{2}}=0 \\
\end{align}\]
Now, on solving this quadratic equation in cos, we get
\[\begin{align}
  & \cos \theta =\dfrac{-2\pm \sqrt{{{2}^{2}}-4(1+{{k}^{2}})(1-{{k}^{2}})}}{2\left( 1+{{k}^{2}} \right)} \\
 & \cos \theta =\dfrac{-2\pm \sqrt{4-4(1+{{k}^{2}})(1-{{k}^{2}})}}{2\left( 1+{{k}^{2}} \right)} \\
 & \cos \theta =\dfrac{-2\pm 2\sqrt{1-(1+{{k}^{2}})(1-{{k}^{2}})}}{2\left( 1+{{k}^{2}} \right)} \\
 & \cos \theta =\dfrac{-1\pm \sqrt{1-1+{{k}^{4}}}}{\left( 1+{{k}^{2}} \right)} \\
 & \cos \theta =\dfrac{-1\pm \sqrt{{{k}^{4}}}}{\left( 1+{{k}^{2}} \right)} \\
 & \cos \theta =\dfrac{-1\pm {{k}^{2}}}{\left( 1+{{k}^{2}} \right)} \\
 & \cos \theta =\dfrac{{{k}^{2}}-1}{{{k}^{2}}+1},-1 \\
\end{align}\]
Now, as the result is equal to the expression mentioned in the question, so, the answer is true.

Note: -The students can make an error if they don’t know about the properties and the relations that are mentioned in the hint as follows
\[\begin{align}
  & \tan x=\dfrac{\sin x}{\cos x} \\
 & \cot x=\dfrac{\cos x}{\sin x} \\
 & \cos ecx=\dfrac{1}{\sin x} \\
 & \sec x=\dfrac{1}{\cos x} \\
\end{align}\]
Without knowing these relations one could not get to the right answer.
Another important formula for solving the question is the quadratic formula which is used to find the roots of a quadratic equation.