If \(\frac{{\sin x + \cos x}}{{\sin x - \cos x}} = \frac{6}{5}\) , then the value of \(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\) is:
Solving Trigonometric Equations: Finding tan x
The problem asks us to find the value of an expression involving \(\tan^2 x\) given an equation relating \(\sin x\) and \(\cos x\). The first step is to use the given equation to find the value of \(\tan x\).
The given equation is:
\[ \frac{{\sin x + \cos x}}{{\sin x - \cos x}} = \frac{6}{5} \]
To find \(\tan x\), we can manipulate this equation. A common technique when dealing with \(\sin x\) and \(\cos x\) in a fraction is to divide the numerator and the denominator by \(\cos x\). This works as long as \(\cos x \neq 0\).
Dividing both the numerator and the denominator of the left side by \(\cos x\):
\[ \frac{{\frac{\sin x}{\cos x} + \frac{\cos x}{\cos x}}}{{\frac{\sin x}{\cos x} - \frac{\cos x}{\cos x}}} = \frac{6}{5} \]
This simplifies to:
\[ \frac{{\tan x + 1}}{{\tan x - 1}} = \frac{6}{5} \]
Now, we can solve for \(\tan x\) by cross-multiplying:
\[ 5(\tan x + 1) = 6(\tan x - 1) \]
Distribute the numbers on both sides:
\[ 5\tan x + 5 = 6\tan x - 6 \]
Gather the \(\tan x\) terms on one side and the constant terms on the other side:
\[ 5 + 6 = 6\tan x - 5\tan x \]
Simplify both sides:
\[ 11 = \tan x \]
So, the value of \(\tan x\) is 11.
Evaluating the Expression \(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\)
Now that we have \(\tan x = 11\), we can find \(\tan^2 x\):
\[ \tan^2 x = (11)^2 = 121 \]
The expression we need to evaluate is \(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\). Substitute the value of \(\tan^2 x\) into this expression:
\[ \frac{{121 + 1}}{{121 - 1}} \]
Perform the addition and subtraction in the numerator and the denominator:
\[ \frac{{122}}{{120}} \]
This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 2:
\[ \frac{{122 \div 2}}{{120 \div 2}} = \frac{61}{60} \]
Thus, the value of the expression \(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\) is \(\frac{61}{60}\).
Comparing with Options
Let's compare our calculated value with the given options:
| Option | Value | Matches Calculation? |
|---|---|---|
| 1 | \(\frac{{35}}{{61}}\) | No |
| 2 | \(\frac{{61}}{{60}}\) | Yes |
| 3 | \(\frac{{60}}{{61}}\) | No |
| 4 | \(\frac{{61}}{{35}}\) | No |
Our calculated value, \(\frac{61}{60}\), matches Option 2.
Step-by-Step Solution Summary
- Start with the given equation \(\frac{{\sin x + \cos x}}{{\sin x - \cos x}} = \frac{6}{5}\).
- Divide numerator and denominator by \(\cos x\) to get \(\frac{{\tan x + 1}}{{\tan x - 1}} = \frac{6}{5}\).
- Cross-multiply and solve for \(\tan x\). We found \(\tan x = 11\).
- Calculate \(\tan^2 x\), which is \(11^2 = 121\).
- Substitute the value of \(\tan^2 x\) into the expression \(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\).
- Evaluate the expression: \(\frac{{121 + 1}}{{121 - 1}} = \frac{{122}}{{120}}\).
- Simplify the fraction: \(\frac{{122}}{{120}} = \frac{61}{60}\).
Revision Table: Key Trigonometric Identities
Understanding basic trigonometric identities is crucial for solving problems like this.
| Identity | Description |
|---|---|
| \(\tan x = \frac{{\sin x}}{{\cos x}}\) | Definition of tangent in terms of sine and cosine. |
| \({{\sec }^2}\theta - {{\tan }^2}\theta = 1\) | Pythagorean identity related to tangent and secant. Can be rearranged as \({{\sec }^2}\theta = 1 + {{\tan }^2}\theta\). |
| \({{\csc }^2}\theta - {{\cot }^2}\theta = 1\) | Pythagorean identity related to cosecant and cotangent. |
Additional Information: Alternative Approach using Componendo and Dividendo
For the equation \(\frac{{\sin x + \cos x}}{{\sin x - \cos x}} = \frac{6}{5}\), we could also use the componendo and dividendo rule. If \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\).
Let \(a = \sin x + \cos x\), \(b = \sin x - \cos x\), \(c = 6\), \(d = 5\).
Applying the rule:
\[ \frac{{(\sin x + \cos x) + (\sin x - \cos x)}}{{(\sin x + \cos x) - (\sin x - \cos x)}} = \frac{{6 + 5}}{{6 - 5}} \]
Simplify the numerator and denominator on the left side:
- Numerator: \( \sin x + \cos x + \sin x - \cos x = 2\sin x \)
- Denominator: \( \sin x + \cos x - \sin x + \cos x = 2\cos x \)
Simplify the right side:
- Numerator: \( 6 + 5 = 11 \)
- Denominator: \( 6 - 5 = 1 \)
So the equation becomes:
\[ \frac{{2\sin x}}{{2\cos x}} = \frac{{11}}{1} \]
\[ \frac{{\sin x}}{{\cos x}} = 11 \]
\[ \tan x = 11 \]
This confirms the value of \(\tan x\) found earlier. Both methods lead to the same result for \(\tan x\).
The expression \(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\) can also be related to other trigonometric functions. Recall the identity \({{\sec }^2}\theta = 1 + {{\tan }^2}\theta\). The numerator is \(1 + {{\tan }^2}x = {{\sec }^2}x\). The denominator is \({{\tan }^2}x - 1\).
So the expression is \(\frac{{{{\sec }^2}x}}{{{{\tan }^2}x - 1}}\). Substituting \(\tan x = 11\):
Numerator: \({{\sec }^2}x = 1 + {{\tan }^2}x = 1 + 121 = 122\)
Denominator: \({{\tan }^2}x - 1 = 121 - 1 = 120\)
The fraction is \(\frac{122}{120} = \frac{61}{60}\), which is consistent with our previous calculation.
Important Questions from Trigonometric Functions
If A = cos2θ + sin4θ then for all values of θ is :
The minimum value of 4 cosθ + 3 is
What is the value of \(? = \frac{{ta{n^2}{{60}^0} - 2si{n^2}{{45}^0}}}{{cos{{24}^0}cos{{37}^0}coses{{53}^0}cos{{60}^0}cosec{{66}^0} + si{n^2}{{60}^0}}}\)
If Y = tan35°, then the value of (2tan55° + cot55°) is :
If -sin θ + cosec θ = 6, then what is the value of sin θ + cosec θ?
