If Tan A-12-5- then the value of sin A-cos A - secA is-

The correct option is B 17/5 ( sin A + cos A ) × sec A = tan A + 1 = 12/5 + 1 = 17/5 ...

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Values of Trigonometric Ratios

If Tan A=12/5...

Question

If TanA = $\frac{12}{5}$ , then the value of (sinA+cosA) $\times$ secA is:

13/5

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17/5

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14/5

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Solution

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\frac{t a n A + s e c A - 1}{t a n A - s e c A + 1} = \frac{1 + s i n A}{c o s A}

tan A = \frac{5}{12}

(sin A + cos A) \times sec A

\frac{1 + s i n A}{1 - s i n A} = s e c A - t a n A

\frac{s i n A - 2 s i n^{2} A}{2 c o s^{3} A - c o s A} = t a n A

secA - t a n A = \frac{cos A}{1 + sin A}

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