Trigonometric Expression Simplification
The problem asks us to find the value of the expression $\frac{s^2-1}{s^2+1}$ given that $s = \sec\theta + \tan\theta$. We need to use trigonometric identities to simplify this expression.
Key Trigonometric Identity
We will use the fundamental Pythagorean identity involving secant and tangent:
$ \sec^2\theta - \tan^2\theta = 1 $This identity can be factored as a difference of squares:
$ (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1 $Expressing $\sec\theta - \tan\theta$ in terms of $s$
We are given $s = \sec\theta + \tan\theta$. Substituting this into the factored identity:
$ (\sec\theta - \tan\theta) \times s = 1 $Solving for $(\sec\theta - \tan\theta)$, we get:
$ \sec\theta - \tan\theta = \frac{1}{s} $Finding Expressions for $s^2 - 1$ and $s^2 + 1$
We now have a system of two equations:
- $\sec\theta + \tan\theta = s$
- $\sec\theta - \tan\theta = \frac{1}{s}$
Let's subtract the second equation from the first:
$ (\sec\theta + \tan\theta) - (\sec\theta - \tan\theta) = s - \frac{1}{s} $ $ 2\tan\theta = \frac{s^2 - 1}{s} $Now, let's add the two equations:
$ (\sec\theta + \tan\theta) + (\sec\theta - \tan\theta) = s + \frac{1}{s} $ $ 2\sec\theta = \frac{s^2 + 1}{s} $Calculating the Final Expression
We need to find the value of $\frac{s^2-1}{s^2+1}$. We can substitute the expressions we found for $2\tan\theta$ and $2\sec\theta$ divided by $2s$:
From $2\tan\theta = \frac{s^2 - 1}{s}$, we have $\frac{s^2 - 1}{2s} = \tan\theta$.
From $2\sec\theta = \frac{s^2 + 1}{s}$, we have $\frac{s^2 + 1}{2s} = \sec\theta$.
Now, let's consider the expression $\frac{s^2-1}{s^2+1}$:
$ \frac{s^2-1}{s^2+1} = \frac{2s \times \tan\theta}{2s \times \sec\theta} $Assuming $s \neq 0$ and $\sec\theta \neq 0$, we can simplify this to:
$ \frac{s^2-1}{s^2+1} = \frac{\tan\theta}{\sec\theta} $Recall the definitions of $\tan\theta$ and $\sec\theta$ in terms of $\sin\theta$ and $\cos\theta$:
$ \tan\theta = \frac{\sin\theta}{\cos\theta} $ $ \sec\theta = \frac{1}{\cos\theta} $Substitute these into the expression:
$ \frac{\tan\theta}{\sec\theta} = \frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}} $Simplifying the fraction:
$ \frac{\sin\theta}{\cos\theta} \times \frac{\cos\theta}{1} = \sin\theta $Thus, the value of the expression $\frac{s^2-1}{s^2+1}$ is $\sin\theta$.
Conclusion
Using the trigonometric identity $\sec^2\theta - \tan^2\theta = 1$ and algebraic manipulation, we found that if $s = \sec\theta + \tan\theta$, then $\frac{s^2-1}{s^2+1} = \sin\theta$.
Important Questions from Trigonometric Identities
- Find the value of $\frac{\sin A - 2\sin^3 A}{2\cos^3 A - \cos A} + \sqrt{\text{cosec}^2 A - 1}$.
- Solve the following:
$(\text{cosec}23^{\circ}-\text{sin}23^{\circ}) (\text{tan}23^{\circ} + \text{cot}23^{\circ}) (\text{sec}23^{\circ}-\text{cos}23^{\circ})$ If $M = \sqrt{\frac{1+\cos A}{1-\cos A}} - \text{cosec} A - \cot A$, and A is an acute angle, then value of M is
- If
$x = a \text{ cosec}A \text{ sin} B,$
$y = b \text{ cosec}A \text{ cos} B,$ and
$z = c \text{ cot} A,$
then, what is the value of $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2}$? Simplify: \( sec^{2} \phi cot^{2 } \phi - \frac{1}{ tan^{2} \phi } +2( \frac{1}{ cosec^{2} \phi }+ \frac{1}{ sec^{2} \phi } )\)
