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Question

If 13tan A + 13 = 17secA, 0 < x < 90; then find sin A - cos A.


  1. 7/19

  2. 5/17

  3. 7/13

  4. 7/10


Solution

The correct option is C

7/13


13tan A + 13 = 17secA

tan A + 1 = 17secA/13

(sin A/cos A) + 1 = 17/13cos A

sin A + cos A = 17/13

Squaring sinA + cosA = 17/13 on both sides,

sin2A + cos2A + 2sinA cosA = 289/169

sinAcosA = 60/169

(sinA - cosA)2 = (sinA + cosA)2 - 4sinAcosA

sinA - cosA = 7/13

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