
If $\tan \dfrac{\theta }{2}=t$, then $\dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}$ is equal to
1. $\cos \theta $
2. $\sin \theta $
3. $\sec \theta $
4. $\cos 2\theta $
Answer
505.8k+ views
Hint: To solve this question we will use the trigonometric formula related to the tangent function. We will substitute the value of $t$ given in the question in the expression $\dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}$ and simplify the obtained expression to get the desired answer.
Formula Used:
$\dfrac{\left( 1-{{\tan }^{2}}\theta \right)}{\left( 1+{{\tan }^{2}}\theta \right)}=\cos 2\theta $.
Complete step-by-step solution:
We have been given that $\tan \dfrac{\theta }{2}=t$.
We have to find the value of $\dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}$.
Now, let us consider the expression $\dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}..........(i)$.
Now, as given in the question $\tan \dfrac{\theta }{2}=t$, substituting the value of $t$ in equation (i) we will get
$\Rightarrow \dfrac{\left( 1-{{\tan }^{2}}\dfrac{\theta }{2} \right)}{\left( 1+{{\tan }^{2}}\dfrac{\theta }{2} \right)}$
Now, we know that $\dfrac{\left( 1-{{\tan }^{2}}\theta \right)}{\left( 1+{{\tan }^{2}}\theta \right)}=\cos 2\theta $
Now, substituting the value in the above obtained equation we will get
$\Rightarrow \cos 2\left( \dfrac{\theta }{2} \right)$
Now, simplifying the above obtained equation we will get
$\Rightarrow \cos \theta $
Hence we get the value of $\dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}$ as $\cos \theta $.
Option A is the correct answer.
Note: To solve such types of questions students must have knowledge of trigonometric functions, ratios and identities. Students must know the relation between the ratios so that they easily convert one ratio into another form to solve the questions. The possibility of mistakes while solving the question is applying the formula. Students may write it as $\cos 2\theta $ and choose the option D which is an incorrect solution. So be careful while solving as $\dfrac{\theta }{2}$ is an angle so we need to write it as it is and then solve further.
Formula Used:
$\dfrac{\left( 1-{{\tan }^{2}}\theta \right)}{\left( 1+{{\tan }^{2}}\theta \right)}=\cos 2\theta $.
Complete step-by-step solution:
We have been given that $\tan \dfrac{\theta }{2}=t$.
We have to find the value of $\dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}$.
Now, let us consider the expression $\dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}..........(i)$.
Now, as given in the question $\tan \dfrac{\theta }{2}=t$, substituting the value of $t$ in equation (i) we will get
$\Rightarrow \dfrac{\left( 1-{{\tan }^{2}}\dfrac{\theta }{2} \right)}{\left( 1+{{\tan }^{2}}\dfrac{\theta }{2} \right)}$
Now, we know that $\dfrac{\left( 1-{{\tan }^{2}}\theta \right)}{\left( 1+{{\tan }^{2}}\theta \right)}=\cos 2\theta $
Now, substituting the value in the above obtained equation we will get
$\Rightarrow \cos 2\left( \dfrac{\theta }{2} \right)$
Now, simplifying the above obtained equation we will get
$\Rightarrow \cos \theta $
Hence we get the value of $\dfrac{\left( 1-{{t}^{2}} \right)}{\left( 1+{{t}^{2}} \right)}$ as $\cos \theta $.
Option A is the correct answer.
Note: To solve such types of questions students must have knowledge of trigonometric functions, ratios and identities. Students must know the relation between the ratios so that they easily convert one ratio into another form to solve the questions. The possibility of mistakes while solving the question is applying the formula. Students may write it as $\cos 2\theta $ and choose the option D which is an incorrect solution. So be careful while solving as $\dfrac{\theta }{2}$ is an angle so we need to write it as it is and then solve further.
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