8 Tan A = 15 then Sin A – Cos A =A) 7/3B) 7/17C) 9/14D) 3/7

1.	8 Tan A = 15 then Sin A – Cos A =A) 7/3B) 7/17C) 9/14D) 3/7
Answer» Correct option is: B) \(\frac{7}{17}\) Given that 8 tan A = 15 = tan A = \(\frac {15}8\) \(\because\) \(sec^2 A = 1 + tan^2 A\) = \(1 + (\frac {15}{8})^2\) = 1 + \(\frac {225}{64}\) = \(\frac {64 + 225}{64} = \frac {289}{64}\) = \((\frac {17}{8})^2\) \(\therefore\) sec A = \(\frac {17}8\) \(\therefore\) cos A = \(\frac 1 {sec \, A} = \frac 1{\frac {17}8} = \frac 8{17}\) \(\because\) tan A = \(\frac {sin\, A}{cos \, A}\) = sin A = cos A. tan A = \(\frac 8 {17} \times \frac {15}8 = \frac {15}{17}\) Now, sin A - cos A = \( \frac {15}{17} - \frac 8 {17} = \frac {15-8}{17} = \frac 7{17}\) Correct option is: B) \(\frac{7}{17}\)

Answer»

Correct option is: B) \(\frac{7}{17}\)

Given that

8 tan A = 15

= tan A = \(\frac {15}8\)

\(\because\) \(sec^2 A = 1 + tan^2 A\)

= \(1 + (\frac {15}{8})^2\)

= 1 + \(\frac {225}{64}\)

= \(\frac {64 + 225}{64} = \frac {289}{64}\)

= \((\frac {17}{8})^2\)

\(\therefore\) sec A = \(\frac {17}8\)

\(\therefore\) cos A = \(\frac 1 {sec \, A} = \frac 1{\frac {17}8} = \frac 8{17}\)

\(\because\) tan A = \(\frac {sin\, A}{cos \, A}\)

= sin A = cos A. tan A = \(\frac 8 {17} \times \frac {15}8 = \frac {15}{17}\)

Now, sin A - cos A = \( \frac {15}{17} - \frac 8 {17} = \frac {15-8}{17} = \frac 7{17}\)

Correct option is: B) \(\frac{7}{17}\)

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