If `(sintheta+costheta)/(sintheta-costheta)=(5)/(4)` , the value of `(tan^(2)theta+1)/(tan^(2)theta-1)` is

Options

A. a)`(25)/(16)`
B. b)`(41)/(9)`
C. c)`(41)/(40)`
D. d)`(40)/(41)`

Answer

To solve the problem, we start with the given equation:

\[ \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} = \frac{5}{4} \]

**Step 1: Cross-multiply to eliminate the fraction.**

\[ 4(\sin \theta + \cos \theta) = 5(\sin \theta - \cos \theta) \]

**Step 2: Expand both sides.**

\[ 4\sin \theta + 4\cos \theta = 5\sin \theta - 5\cos \theta \]

**Step 3: Rearrange the equation to isolate terms involving \(\sin \theta\) and \(\cos \theta\).**

\[ 4\sin \theta + 4\cos \theta - 5\sin \theta + 5\cos \theta = 0 \]

This simplifies to:

\[ - \sin \theta + 9\cos \theta = 0 \]

**Step 4: Solve for \(\tan \theta\).**

Rearranging gives:

\[ \sin \theta = 9\cos \theta \]

Dividing both sides by \(\cos \theta\) (assuming \(\cos \theta \neq 0\)):

\[ \tan \theta = 9 \]

**Step 5: Find \(\tan^2 \theta\).**

\[ \tan^2 \theta = 9^2 = 81 \]

**Step 6: Substitute \(\tan^2 \theta\) into the expression \(\frac{\tan^2 \theta + 1}{\tan^2 \theta - 1}\).**

\[ \frac{\tan^2 \theta + 1}{\tan^2 \theta - 1} = \frac{81 + 1}{81 - 1} = \frac{82}{80} \]

**Step 7: Simplify the fraction.**

\[ \frac{82}{80} = \frac{41}{40} \]

Thus, the final answer is:

\[ \frac{41}{40} \]

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