If \(\frac{{\sin x + \cos x}}{{\sin x - \cos x}} = \frac{6}{5}\), the

1.	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:1. \(\frac{{35}}{{61}}\)2. \(\frac{{61}}{{60}}\)3. \(\frac{{60}}{{61}}\)4. \(\frac{{61}}{{35}}\)
Answer» Correct Answer - Option 2 : \(\frac{{61}}{{60}}\) Given: \(\frac{{\sin x + \cos x}}{{\sin x - \cos x}} = \frac{6}{5}\) Concept Used: Componendo & dividendo a/b = c/d then (a + b)/(a - b) =(c + d)/(c - d) Calculation: \(\frac{{\sin x + \cos x}}{{\sin x - \cos x}} = \frac{6}{5}\) By using componendo & dividendo (sinx + cosx + sinx - cos x)/(sinx + cosx - sinx + cosx) = (6 + 5)/(6 - 5) (2sinx/2cosx) = (sinx/cosx) = 11/1 tan²x = (sin²x/cos²x) = 11²/1² = 121 tan²x + 1 = 121 +1 =122 tan²x - 1 = 121 - 1 = 120 \(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\) = 122/120 = 61/60 ∴ The value of \(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\) is 61/60

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:1. \(\frac{{35}}{{61}}\)2. \(\frac{{61}}{{60}}\)3. \(\frac{{60}}{{61}}\)4. \(\frac{{61}}{{35}}\)

Answer» Correct Answer - Option 2 : \(\frac{{61}}{{60}}\)

Given:

\(\frac{{\sin x + \cos x}}{{\sin x - \cos x}} = \frac{6}{5}\)

Concept Used:

Componendo & dividendo

a/b = c/d

then (a + b)/(a - b) =(c + d)/(c - d)

Calculation:

\(\frac{{\sin x + \cos x}}{{\sin x - \cos x}} = \frac{6}{5}\)

By using componendo & dividendo

(sinx + cosx + sinx - cos x)/(sinx + cosx - sinx + cosx) = (6 + 5)/(6 - 5)

(2sinx/2cosx) = (sinx/cosx) = 11/1

tan²x = (sin²x/cos²x) = 11²/1² = 121

tan²x + 1 = 121 +1 =122

tan²x - 1 = 121 - 1 = 120

\(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\) = 122/120 = 61/60

∴ The value of \(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\) is 61/60