Question

درة
1. In a right angled triangle ABC, angleA is acute, angleB = 90° and tan A =3/5
Find
1) sin A
ii) cosecA - Cota​

Answer 1

sinA = 3/√34, cosecA-cotA = (√34+5)/3

Step-by-step explanation:

We are given that, tanA = 3/5. This means:

⇒ Perpendicular/Base = 3/5

Let the Perpendicular be 3k and base be 5k. Now, let us find the hypotenuse of the right angled triangle by Pythagoras Theorem.

(Pythagoras theorem: In a right-angled triangle, the square of hypotenuse is equal to the sum of square of other two sides.)

On finding Hypotenuse, we get:

⇒ (Hypotenuse)² = (Perpendicular)²+(Base)²

⇒ (Hypotenuse)² = (3k)²+(5k)²

⇒ (Hypotenuse)² = 9k²+25k²

⇒ (Hypotenuse)² = 34k²

⇒ Hypotenuse = √34k²

⇒ Hypotenuse = √34 k

Therefore, the Hypotenuse of the triangle is √34 k. Now, we know that:

1) sinA = Perpendicular/Hypotenuse

⇒ sinA = 3k/√34 k

⇒ sinA = 3/√34

Hence, sinA = 3/√34

2) cosecA = Hypotenuse/Perpendicular

⇒ cosecA = √34 k/3k

⇒ cosecA = √34/3

We also know that:

→ CotA = 1/tanA

→ CotA = 1/(3/5)

→ CotA = 5/3

Hence, let's come to the question:

⇒ cosecA - cotA = √34/3+5/3

⇒ cosecA - cotA = (√34+5)/3

Hence, the value of cosecA - cotA = (√34+5)/3.