Question

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

Answer 1

Hint: 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 \[\dfrac{{2P}}{{{P^2} + 1}}\].