Question

If $\sec \theta +\tan \theta =p$, then $\tan \theta$ is equal to
$1)\text{ }2p/\left( {{p}^{2}}-1 \right)$
$2)\text{ }\left( {{p}^{2}}-1 \right)/2p$
$3)\text{ }\left( {{p}^{2}}+1 \right)/2p$
$4)\text{ }2p/\left( {{p}^{2}}+1 \right)$

Answer 1

In this question we have been given a trigonometric equation and based on the equation we have to find the value of $\tan \theta$. We will solve this question by using the trigonometric identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ and expand it. We will also use the identity that $\sin \theta =\sqrt{\left( 1-{{\cos }^{2}}\theta \right)}$ and then use the identity $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and get the required solution.

Complete step-by-step solution:
We have the expression given to us as:
$\Rightarrow \sec \theta +\tan \theta =p\to \left( 1 \right)$
And based on this we have to find the value of $\tan \theta$.
We know the trigonometric identity that ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$.
On using the expansion formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we get:
$\Rightarrow \left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta \right)=1$
On rearranging, we get:
$\Rightarrow \left( \sec \theta -\tan \theta \right)=\dfrac{1}{\left( \sec \theta +\tan \theta \right)}$
From equation $\left( 1 \right)$, we get:
$\Rightarrow \left( \sec \theta -\tan \theta \right)=\dfrac{1}{p}\to \left( 2 \right)$
Now on adding $\left( 1 \right)$ and $\left( 2 \right)$, we get:
$\Rightarrow 2\sec \theta =p+\dfrac{1}{p}$
On taking the lowest common multiple in the fractions, we get:
$\Rightarrow 2\sec \theta =\dfrac{{{p}^{2}}+1}{p}$
On transferring $2$ from the left-hand side to the right-hand side, we get:
$\Rightarrow \sec \theta =\dfrac{{{p}^{2}}+1}{2p}$
Now we know that $\cos \theta =\dfrac{1}{\sec \theta }$ therefore, we can write:
$\Rightarrow \cos \theta =\dfrac{1}{\left( {{p}^{2}}+1/2p \right)}$
We get the expression as:
$\Rightarrow \cos \theta =\dfrac{2p}{{{p}^{2}}+1}$
Now we know that $\sin \theta =\sqrt{\left( 1-{{\cos }^{2}}\theta \right)}$ on substituting the value of $\theta$, we get:
$\Rightarrow \sin \theta =\sqrt{\left( 1-{{\left( \dfrac{2p}{{{p}^{2}}+1} \right)}^{2}} \right)}$
On expanding the terms, we get:
$\Rightarrow \sin \theta =\sqrt{\left( 1-\dfrac{4{{p}^{2}}}{{{\left( {{p}^{2}}+1 \right)}^{2}}} \right)}$
On taking the lowest common multiple, we get:
$\Rightarrow \sin \theta =\sqrt{\left( \dfrac{{{\left( {{p}^{2}}+1 \right)}^{2}}-4{{p}^{2}}}{{{\left( {{p}^{2}}+1 \right)}^{2}}} \right)}$
On using the expansion formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get:
$\Rightarrow \sin \theta =\sqrt{\left( \dfrac{{{p}^{4}}+2{{p}^{2}}+1-4{{p}^{2}}}{{{\left( {{p}^{2}}+1 \right)}^{2}}} \right)}$
On simplifying, we get:
$\Rightarrow \sin \theta =\sqrt{\left( \dfrac{{{p}^{4}}-2{{p}^{2}}+1}{{{\left( {{p}^{2}}+1 \right)}^{2}}} \right)}$
Now we can see that ${{p}^{4}}-2{{p}^{2}}+1$ is the expansion of ${{\left( {{p}^{2}}-1 \right)}^{2}}$ therefore, on substituting, we get:
$\Rightarrow \sin \theta =\sqrt{\left( \dfrac{{{\left( {{p}^{2}}-1 \right)}^{2}}}{{{\left( {{p}^{2}}+1 \right)}^{2}}} \right)}$
On taking the square root, we get:
$\Rightarrow \sin \theta =\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}$
Now we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, on substituting the values, we get:
$\Rightarrow \tan \theta =\dfrac{\dfrac{{{p}^{2}}-1}{{{p}^{2}}+1}}{\dfrac{2p}{{{p}^{2}}+1}}$
On simplifying, we get:
$\Rightarrow \tan \theta =\dfrac{{{p}^{2}}-1}{2p}$, which is the required value.
Therefore, the correct answer is option $\left( 2 \right)$.

Note: In these types of questions, the trigonometric identities should be remembered to solve the question. The various trigonometric identities and formulae should be remembered while doing these types of sums. The various Pythagorean identities should also be remembered while doing these types of questions.