Question

If $\sec \theta + \tan \theta = m$, how do you show that $\dfrac{{{m^2} - 1}}{{{m^2} + 1}} = \sin \theta$?

Answer 1

This question is from the topic of trigonometric identities. In this question we need to prove $\dfrac{{{m^2} - 1}}{{{m^2} + 1}} = \sin \theta$ if $\sec \theta + \tan \theta = m$, to prove this we will first put given value of $m$ in the equation and simplify it by trigonometric identities and relation between trigonometric functions.

Complete step by step solution: Let us try to solve this question in which we are asked to prove that $\dfrac{{{m^2} - 1}}{{{m^2} + 1}} = \sin \theta$ where$\sec \theta + \tan \theta = m$. To prove this we will first put the value of $m$ in the equation. After putting the value of$m$, we use trigonometric identities such as ${\sec ^2}x - {\tan ^2}x = 1$ to simplify it and finally to reach $\sin \theta$. So, let’s try to prove.

To prove: $\dfrac{{{m^2} - 1}}{{{m^2} + 1}} = \sin \theta$ given that $\sec \theta + \tan \theta = m$.
Proof: We have,
$\dfrac{{{m^2} - 1}}{{{m^2} + 1}} = \sin \theta$ $(1)$
Putting the value of $m$in the equation$(1)$, we get
$\dfrac{{{{(\sec \theta + \tan \theta )}^2} - 1}}{{{{(\sec \theta + \tan \theta )}^2} + 1}} = \sin \theta$ $(2)$
Now, applying algebraic identity ${(a + b)^2} = {a^2} + 2ab + {b^2}$ to the equation$(2)$, we get
$\dfrac{{{{\sec }^2}\theta + 2\sec \theta \tan \theta + {{\tan }^2}\theta - 1}}{{{{\sec }^2}\theta + 2\sec \theta \tan \theta + {{\tan }^2}\theta + 1}} = \sin \theta$ $(3)$
Now, we will apply trigonometric identities in equation $(3)$to simplify it.
Now by using trigonometric identity${\sec ^2}x - {\tan ^2}x = 1$, we have
$\dfrac{{{{\sec }^2}\theta + 2\sec \theta \tan \theta + {{\tan }^2}\theta - ({{\sec }^2}\theta - {{\tan }^2}\theta )}}{{{{\sec }^2}\theta + 2\sec \theta \tan \theta + {{\tan }^2}\theta + ({{\sec }^2}\theta \- {{\tan }^2}\theta )}} = \sin \theta$ $(4)$
In the numerator ${\sec ^2}\theta$cancel each other. Similarly, in denominator ${\sec ^2}\theta$terms
cancel each other. We get,

2\sec \theta \tan \theta }} = \sin \theta $$ $(5)$ Taking $2$common from equation $(5)$ both numerator and denominator L.H.S, we get $$\dfrac{{\sec \theta \tan \theta + {{\tan }^2}\theta }}{{{{\sec }^2}\theta + \sec \theta \tan \theta }} = \sin \theta $$ $(6)$ As we know $\sec \theta = \dfrac{1}{{\cos \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ in equation$(6)$, we get $\dfrac{{\dfrac{1}{{\cos \theta }} \cdot \dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{\dfrac{1}{{{{\cos }^2}\theta }} + \dfrac{1}{{\cos \theta }} \cdot \dfrac{{\sin \theta }}{{\cos \theta }}}} = \sin \theta $ $(7)$ Now, simplifying equation $(7)$by performing fraction addition both in numerator and denominator of equation $(7)$L.H.S, we get $\dfrac{{\dfrac{{\sin \theta + {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{\dfrac{{1 \+ \sin \theta }}{{{{\cos }^2}\theta }}}} = \sin \theta $ $(8)$ Now after cancelling ${\cos ^2}\theta $ from the L.H.S of equation$(8)$, we get $\dfrac{{\sin \theta (1 + \sin \theta )}}{{1 + \sin \theta }} = \sin \theta $ $(9)$ Now, by cancelling $1 + \sin \theta $ in the L.H.S of equation $(9)$, we get our required result. $\sin \theta = \sin \theta $ Since we have shown L.H.S equal to R.H.S. Hence proved, if $\sec \theta + \tan \theta = m$then$\dfrac{{{m^2} - 1}}{{{m^2} + 1}} = \sin \theta $. **Note:** While solving these types of questions in which we have to prove trigonometric equations, we will start with the L.H.S of equation to derive R.H.S from it. To prove this question we only requires knowledge of basic trigonometric identities such as ${\sin ^2}x + {\cos ^2}x = 1$,${\sec ^2}x - {\tan ^2}x = 1$