In a right-angled triangle XYZ right-angled at Y. if XY = $$2\sqrt 6 $$ and XZ - YZ = 2, then secX + tanX is:
A. $$\frac{1}{{\sqrt 6 }}$$
B. $$\sqrt 6 $$
C. $$2\sqrt 6 $$
D. $$\frac{{\sqrt 6 }}{2}$$
Answer: Option B
Solution (By Examveda Team)
XZ - YZ = 2
h - P = 2 . . . . . . . (i)
h2 = (2√6)2 + P2
h2 - P2 = (2√6)2
(h - P)(h + P) = 4 × 6
(2)(h + P) = 24
h + P = 12 . . . . . . . (ii)
Adding equation (i) and (ii)
2h = 14
h = 7, P = 5
$$\eqalign{ & \sec {\text{X}} + \tan {\text{X}} \cr & = \frac{{\text{h}}}{{{\text{XY}}}} + \frac{{\text{P}}}{{{\text{XY}}}} \cr & = \frac{7}{{2\sqrt 6 }} + \frac{5}{{2\sqrt 6 }} \cr & = \frac{{12}}{{2\sqrt 6 }} \cr & = \frac{6}{{\sqrt 6 }} \cr & = \sqrt 6 \cr} $$
This Question Belongs to Arithmetic Ability >> Circular Measurement Of Angle
Join The Discussion