Question

if tanA = m/n, then find sinA and cosA​

Answer 1

Step-by-step explanation:

Given :-

Tan A = m / n

To find :-

Find Sin A and Cos A ?

Solution:-

Given that:

Tan A = m / n

On squaring both sides then

=> Tan² A = (m/n)²

=> Tan² A = m²/n²

On adding 1 both sides then

=> 1 +Tan² A = 1+(m²/n²)

=> 1+Tan² A = (n²+m²)/n²

=> Sec² A = (m²+n²)/n²

Since , Sec² A - Tan² A = 1

=> Sec A = √[(m²+n²)/n²]

=> Sec A = √(m²+n²) / n

=> 1/Cos A = √(m²+n²) / n

=> Cos A = n/√(m²+n²) -----------(1)

On squaring both sides then

=> Cos² A = [n/√(m²+n²)]²

=> Cos² A = n²/(m²+n²)

On Subtracting above equation from 1 both sides

=> 1 - Cos² A = 1-[n²/(m²+n²)]

=> 1- Cos² A = (m²+n²-n²)/(m²+n²)

=> 1-Cos² A = m²/(m²+n²)

=> Sin² A = m²/(m²+n²)

Since Sin² A + Cos² A = 1

=> Sin A = √[m²/(m²+n²)]

=> Sin A = m/√(m²+n²)------------(2)

Answer:-

The value of Sin A = m/√(m²+n²)

The value of Cos A = n/√(m²+n²)

Used formulae:-

• Sin² A + Cos² A = 1
• Sec² A - Tan² A = 1
• Sec A = 1 / Cos A