Question

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

• A. 7/19
• B. 5/17
• C. 7/10
• D. 7/13

Answer 1

The correct option is D

7/13

13tan A + 13 = 17secA

$\Rightarrow$ tan A + 1 = 17secA/13

$\Rightarrow$ (sin A/cos A) + 1 = 17/13cos A

$\Rightarrow$ sin A + cos A = 17/13

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

sin${}^2$A + cos${}^2$A + 2sinA cosA = 289/169

sinAcosA = 60/169

(sinA - cosA)${}^2$ = (sinA + cosA)${}^2$ - 4sinAcosA

sinA - cosA = 7/13