Question

If Sin A = 3/5 , Prove that Tan A + 1/Cos A = 2

Answer 1

Hyy Dude ♡

➦Draw right-angled triangle ABC such that Angle C is the right angle.

then,

➦sin A = BC/AB = 3/5

➦Let BC = 3 and AB = 5.

by Pythagoras theorem,

• ➨AC² = AB² - BC²

• ➨5² - 3² = 16

• ➨AC = √16

• ➨AC = 4

Tan A = BC/AC = 3/4 and AC/AB = 4/5

Tan A + 1/cos A = 3/4 + 5/4 = 2

Hope it's helps you ;-)

Answer 2

Step-by-step explanation:

Given -

• SinA = 3/5

To Proof -

• TanA + 1/CosA = 2

As we know that :-

SinA = perpendicular/hypotenuse

→ SinA = 3/5

→ perpendicular = 3

And

Hypotenuse = 5

At first we need to calculate its base

As we know that :-

From Pythagoras theorem :-

(hypotenuse)² = (perpendicular)² + (base)²

→ (5)² - (3)² = (base)²

→ base = √25-9

→ base = √16

→ base = 4

Now,

As we know that :-

• TanA = perpendicular/base

→ TanA = 3/4

And

• CosA = base/hypotenuse

→ CosA = 4/5

Now,

• TanA + 1/CosA = 2

→ 3/4 + 1/4/5 = 2

→ 3/4 + 5/4 = 2

→ 3+5/4 = 2

→ 8/4 = 2

→ 2 = 2

LHS = RHS

Hence,

Proved...