Question

If $\sec \theta + \tan \theta = P$, then what is $\cos \theta$ equal to?

Answer 1

First we will convert the given equation into $\sin \theta$ and $\cos \theta$. Then as we know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$, we will form a quadratic equation in $\sin \theta$. Then we will substitute $\sin \theta = t$ to form a quadratic equation in $t$. We will solve this quadratic equation to find $t$ and then we will substitute back $t = \sin \theta$. Then using $\cos \theta = \sqrt {1 - {{\sin }^2}\theta }$ we will find the value of $\cos \theta$.

Complete step by step answer:
Given, $\sec \theta + \tan \theta = P$.
Converting the given equation into $\sin \theta$ and $\cos \theta$, we get
$\Rightarrow \dfrac{1}{{\cos \theta }} + \dfrac{{\sin \theta }}{{\cos \theta }} = P$
On taking the LCM, we get
$\Rightarrow \dfrac{{1 + \sin \theta }}{{\cos \theta }} = P$
Multiplying $\cos \theta$ on both the sides, we get
$\Rightarrow 1 + \sin \theta = P\cos \theta$
Squaring both the sides, we get
$\Rightarrow {\left( {1 + \sin \theta } \right)^2} = {\left( {P\cos \theta } \right)^2}$
On simplifying, we get
$\Rightarrow 1 + 2\sin \theta + {\sin ^2}\theta = {P^2}{\cos ^2}\theta - - - (1)$
As we know that ${\cos ^2}\theta + {\sin ^2}\theta = 1$ i.e., ${\cos ^2}\theta = 1 - {\sin ^2}\theta$.
Using this in $(1)$, we get
$\Rightarrow 1 + 2\sin \theta + {\sin ^2}\theta = {P^2}\left( {1 - {{\sin }^2}\theta } \right)$
$\Rightarrow 1 + 2\sin \theta + {\sin ^2}\theta = {P^2} - {P^2}{\sin ^2}\theta$
On simplifying, we get
$\Rightarrow \left( {{P^2} + 1} \right){\sin ^2}\theta + 2\sin \theta + 1 - {P^2} = 0$
Let $\sin \theta = t$. So, we get
$\Rightarrow \left( {{P^2} + 1} \right){t^2} + 2t + 1 - {P^2} = 0$
As we know that the roots of a quadratic equation $a{x^2} + bx + c = 0$ where $a \ne 0$ is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Using this, we get
$\Rightarrow t = \dfrac{{ - 2 \pm \sqrt {{{\left( 2 \right)}^2} - 4\left( {{P^2} + 1} \right)\left( {1 - {P^2}} \right)} }}{{2\left( {{P^2} + 1} \right)}}$
Using the identity $(a + b)(a - b) = {a^2} - {b^2}$, we get
$\Rightarrow t = \dfrac{{ - 2 \pm \sqrt {4 - 4\left( {1 - {P^4}} \right)} }}{{2\left( {{P^2} + 1} \right)}}$
Taking $4$ common from the root, we get
$\Rightarrow t = \dfrac{{ - 2 \pm 2\sqrt {1 - \left( {1 - {P^4}} \right)} }}{{2\left( {{P^2} + 1} \right)}}$
On simplification, we get
$\Rightarrow t = \dfrac{{ - 2 \pm 2\sqrt {{P^4}} }}{{2\left( {{P^2} + 1} \right)}}$
Rewriting ${P^4}$as ${\left( {{P^2}} \right)^2}$, we get
$\Rightarrow t = \dfrac{{ - 2 \pm 2\sqrt {{{\left( {{P^2}} \right)}^2}} }}{{2\left( {{P^2} + 1} \right)}}$
As $\sqrt {{{\left( a \right)}^2}} = a$. Using this, we get
$\Rightarrow t = \dfrac{{ - 2 \pm 2{P^2}}}{{2\left( {{P^2} + 1} \right)}}$
Cancelling the common term from the numerator and the denominator, we get
$\Rightarrow t = \dfrac{{ - 1 \pm {P^2}}}{{{P^2} + 1}}$
$\Rightarrow t = \dfrac{{ - 1 + {P^2}}}{{{P^2} + 1}}$ or $t = \dfrac{{ - 1 - {P^2}}}{{{P^2} + 1}}$
On rewriting, we get
$\Rightarrow t = \dfrac{{{P^2} - 1}}{{{P^2} + 1}}$ or $t = \dfrac{{ - \left( {{P^2} + 1} \right)}}{{{P^2} + 1}}$
On simplification, we get
$\Rightarrow t = \dfrac{{{P^2} - 1}}{{{P^2} + 1}}$ or $t = - 1$
Substituting $\sin \theta = t$, we get
$\Rightarrow \sin \theta = \dfrac{{{P^2} - 1}}{{{P^2} + 1}}$ or $\sin \theta = - 1$
Using $\cos \theta = \sqrt {1 - {{\sin }^2}\theta }$.
When $\sin \theta = \dfrac{{{P^2} - 1}}{{{P^2} + 1}}$.
$\Rightarrow \cos \theta = \sqrt {1 - {{\left( {\dfrac{{{P^2} - 1}}{{{P^2} + 1}}} \right)}^2}}$
$\Rightarrow \cos \theta = \sqrt {1 - \dfrac{{{P^4} - 2{P^2} + 1}}{{{P^4} + 2{P^2} + 1}}}$
On taking the LCM, we get
$\Rightarrow \cos \theta = \sqrt {\dfrac{{{P^4} + 2{P^2} + 1 - {P^4} + 2{P^2} - 1}}{{{P^4} + 2{P^2} + 1}}}$
$\Rightarrow \cos \theta = \sqrt {\dfrac{{{{\left( {2P} \right)}^2}}}{{{{\left( {{P^2} + 1} \right)}^2}}}}$
On simplifying, we get
$\Rightarrow \cos \theta = \dfrac{{2P}}{{{P^2} + 1}}$
When $\sin \theta = - 1$.
$\Rightarrow \cos \theta = \sqrt {1 - {{\left( { - 1} \right)}^2}}$
$\Rightarrow \cos \theta = 0$
But, $\cos \theta = 0$ and we know that $\tan \theta$ is undefined at all points where $\cos \theta = 0$ i.e., at $\theta = \left( {2n + 1} \right)\dfrac{\pi }{2}$.
$\therefore \cos \theta \ne 0$
Therefore, $\cos \theta$ is equal to $\dfrac{{2P}}{{{P^2} + 1}}$.

Note:
We can also solve this problem by another method.
As we know, ${\sec ^2}\theta - {\tan ^2}\theta = 1$.
$\Rightarrow \left( {\sec \theta - \tan \theta } \right)\left( {\sec \theta + \tan \theta } \right) = 1 - - - (1)$
Given, $\sec \theta + \tan \theta = P$. Using this, we get
$\Rightarrow \left( {\sec \theta - \tan \theta } \right)P = 1$
$\Rightarrow \sec \theta - \tan \theta = \dfrac{1}{P} - - - (2)$
Putting $(2)$ in $(1)$, we get
$\Rightarrow \dfrac{1}{P} \times \left( {\sec \theta + \tan \theta } \right) = 1$
$\Rightarrow \sec \theta + \tan \theta = P - - - (3)$
Adding $(2)$ and $(3)$, we get
$\Rightarrow \sec \theta - \tan \theta + \sec \theta + \tan \theta = \dfrac{1}{P} + P$
$\Rightarrow 2\sec \theta = \dfrac{1}{P} + P$
Taking the LCM on RHS, we get
$\Rightarrow 2\sec \theta = \dfrac{{1 + {P^2}}}{P}$
Dividing both the sides by $2$, we get
$\Rightarrow \sec \theta = \dfrac{{1 + {P^2}}}{{2P}}$
As $\sec \theta = \dfrac{1}{{\cos \theta }}$, we get
$\Rightarrow \dfrac{1}{{\cos \theta }} = \dfrac{{1 + {P^2}}}{{2P}}$
On simplifying, we get
$\Rightarrow \cos \theta = \dfrac{{2P}}{{1 + {P^2}}}$
Therefore, $\cos \theta$ equal to $\dfrac{{2P}}{{{P^2} + 1}}$.