Question

if sinΦ=m/n then find the value of tanΦ+4/cotΦ+1​

Answer 1

$\huge{\pink{Answer}}$



Sinθ=m/n
∴, cosθ=√1-sin²θ=√1-m²/n²=√(n²-m²)/n²=√(n²-m²)/n
∴,tanθ=sinθ/cosθ=m/√(n²-m²)
cotθ=1/tanθ=√(n²-m²)/m
∴, (tanθ+4)/(4cotθ+1)
=[m/√(n²-m²)+4]/[4√(n²-m²)/m+1]
=[{m+4√(n²-m²)}/√(n²-m²)]/[{4√(n²-m²)+m}/m]
={m+4√(n²-m²)}/√(n²-m²)×m/{m+4√(n²-m²}
=m/√(n²-m²)

Answer 2

Answer:then (sinΦ)(1/√2) + (cosΦ)(1/√2) = 1. (sinΦ)(cos(π/4)) + (cosΦ)(sin(π/4)) = 1. sin(Φ+π/4) = 1. Φ + π/4 ... So, tanΦ + cotΦ = 1 + 1 = 2 ... tan∅+cot∅ = (sin2φ + cos2φ) / (sinφ cosφ) = 1 / (sinφ cosφ)

Step-by-step explanation: