Question

If sin A = 3/5, prove that tan A + 1/cos A = 2, if A is an acute angle.

Answer 1

Let's imagine a right-angle triangle ABC in which <c = 90°, BC is the perpendicular, CA is the base and AB is the hypotenuse.

sin A = BC/AB = 3/5

Let, BC = 3 and AB = 5.

By Pythagoras theorem,

AC² = AB²- BC²

      = 5²-3²

      = 25-9

      = 16

∴ AC = √16

         = 4

tan A = BC/AC

         = 3/4

cos A = AC/AB

         = 4/5

∴ tan A + 1/cos A

= 3/4 + 5/4

= 3+5/4

= 8/4

= 2

Hence L.H.S.=R.H.S. (is proved)