Question

If $ \sec \theta -\cos \theta =1 $ then $ {{\tan }^{2}}\dfrac{\theta }{2}= $ .
\[\begin{align}
  & A.\sqrt{5}+2 \\
 & B.\sqrt{5}-2 \\
 & C.2-\sqrt{5} \\
\end{align}\]

Answer 1

Hint: In this question, we are given the value of $ \sec \theta -\cos \theta $ as 1 and we need to find the value of $ {{\tan }^{2}}\dfrac{\theta }{2} $ . For this, we will first convert $ \sec \theta $ into $ \cos \theta $ using the property $ \cos \theta =\dfrac{1}{\sec \theta } $ . After that, we will form a quadratic equation in $ \cos \theta $ solving it will give us value of $ \cos \theta $ . Then we will use the formula $ \cos \theta =\dfrac{1-{{\tan }^{2}}\dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{2}} $ to evaluate $ {{\tan }^{2}}\dfrac{\theta }{2} $ .

Complete step by step answer:
Here we are given the value of $ \sec \theta -\cos \theta $ as 1.
Let us first calculate the value of $ \cos \theta $ using this question. We know that, $ \sec \theta =\dfrac{1}{\cos \theta } $ so putting it in the equation we get,
 $ \dfrac{1}{\cos \theta }-\cos \theta =1 $ .
Taking LCM as $ \cos \theta $ we get, $ \dfrac{1-{{\cos }^{2}}\theta }{\cos \theta }=1 $ .
Cross multiplying we get, $ 1-{{\cos }^{2}}\theta =\cos \theta $ .
Rearranging we get, $ {{\cos }^{2}}\theta +\cos \theta -1=0\cdots \cdots \cdots \left( 1 \right) $ .
This is a quadratic equation. So let us use the quadratic formula to evaluate the value of $ \cos \theta $ .
Quadratic formula for an equation $ a{{x}^{2}}+bx+c $ is given by $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ .
For given equation (1) we have,
\[\begin{align}
  & \cos \theta =\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow \cos \theta =\dfrac{-1\pm \sqrt{1+4}}{2} \\
 & \Rightarrow \cos \theta =\dfrac{-1\pm \sqrt{5}}{2} \\
\end{align}\]
So we get values of $ \cos \theta $ as $ \dfrac{-1+\sqrt{5}}{2}\text{ and }\dfrac{-1-\sqrt{5}}{2} $ .
We know that $ \cos \theta =\dfrac{1-{{\tan }^{2}}\dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{2}} $ .
Putting the value of $ \cos \theta $ as $ \dfrac{-1+\sqrt{5}}{2} $ we get,
 $ \dfrac{-1+\sqrt{5}}{2}=\dfrac{1-{{\tan }^{2}}\dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{2}} $ .
Now let us solve this equation to find the value of $ {{\tan }^{2}}\dfrac{\theta }{2} $ cross multiplying we get,
 $ \left( -1+\sqrt{5} \right)\left( 1+{{\tan }^{2}}\dfrac{\theta }{2} \right)=2\left( 1-{{\tan }^{2}}\dfrac{\theta }{2} \right) $ .
Simplifying we get,
\[-1-{{\tan }^{2}}\dfrac{\theta }{2}+\sqrt{5}+\sqrt{5}{{\tan }^{2}}\dfrac{\theta }{2}=2-2{{\tan }^{2}}\dfrac{\theta }{2}\].
Taking $ {{\tan }^{2}}\dfrac{\theta }{2} $ term on the left side and constant on the other side we get,
\[-{{\tan }^{2}}\dfrac{\theta }{2}+\sqrt{5}{{\tan }^{2}}\dfrac{\theta }{2}+2{{\tan }^{2}}\dfrac{\theta }{2}=2+1-\sqrt{5}\].
Simplifying we get,
\[{{\tan }^{2}}\dfrac{\theta }{2}+\sqrt{5}{{\tan }^{2}}\dfrac{\theta }{2}=3-\sqrt{5}\].
Taking $ {{\tan }^{2}}\dfrac{\theta }{2} $ common from the left side of the equation we get,
\[{{\tan }^{2}}\dfrac{\theta }{2}\left( 1+\sqrt{5} \right)=3-\sqrt{5}\].
Dividing both sides by \[\left( 1+\sqrt{5} \right)\] we get,
\[{{\tan }^{2}}\dfrac{\theta }{2}=\dfrac{3-\sqrt{5}}{1+\sqrt{5}}\].
Since this answer is none of our options so let us rationalize it. Multiplying and dividing the right hand side of the equation by \[1-\sqrt{5}\] we get,
\[{{\tan }^{2}}\dfrac{\theta }{2}=\dfrac{\left( 3-\sqrt{5} \right)\left( 1-\sqrt{5} \right)}{\left( 1+\sqrt{5} \right)\left( 1-\sqrt{5} \right)}\].
Using $ \left( a+b \right)\left( a-b \right) $ in the denominator and simplifying the numerator we get,
\[\begin{align}
  & {{\tan }^{2}}\dfrac{\theta }{2}=\dfrac{3-3\sqrt{5}-\sqrt{5}+\sqrt{5}\sqrt{5}}{1-{{\left( \sqrt{5} \right)}^{2}}} \\
 & \Rightarrow {{\tan }^{2}}\dfrac{\theta }{2}=\dfrac{3-4\sqrt{5}+5}{1-5} \\
 & \Rightarrow {{\tan }^{2}}\dfrac{\theta }{2}=\dfrac{8-4\sqrt{5}}{-4} \\
\end{align}\].
Separating the terms we get,
\[\begin{align}
  & \Rightarrow {{\tan }^{2}}\dfrac{\theta }{2}=\dfrac{-8}{-4}+\dfrac{4\sqrt{5}}{4} \\
 & \Rightarrow {{\tan }^{2}}\dfrac{\theta }{2}=-2+\sqrt{5} \\
 & \Rightarrow {{\tan }^{2}}\dfrac{\theta }{2}=\sqrt{5}-2 \\
\end{align}\].
Hence the value of $ {{\tan }^{2}}\dfrac{\theta }{2} $ is $ \sqrt{5}-2 $ .
Hence option B is the correct answer.

Note:
 Students can also put value of $ \cos \theta =\dfrac{-1-\sqrt{5}}{2} $ and find the value of $ {{\tan }^{2}}\dfrac{\theta }{2} $ but answer will not be in our options. So we did not solve that. If there were no options, students had to find both the values of $ {{\tan }^{2}}\dfrac{\theta }{2} $ . Take care of the sign while solving the sum. In the case of an irrational number in the denominator, always try to rationalize it. Keep in mind all the properties of trigonometry.