Question

If 3 cot ∅ = 4 , then (3 sin ∅ + 4 cos ∅)/(3 sin ∅ - 4 cos ∅) ?​

Answer 1

Explanation:

`\large\underline{\sf{Solution-}}`

`\rm \longmapsto\:3cot\theta = 4`

`\rm\implies \:cot\theta = (4)/(3)`

Now, Consider

`\rm \longmapsto\:(3sin\theta + 4cos\theta )/(3sin\theta - 4cos\theta )`

`\rm \:  =  \: (sin\theta \bigg[3 + 4(cos\theta )/(sin\theta ) \bigg])/(sin\theta \bigg[3 - 4(cos\theta )/(sin\theta ) \bigg])`

`\rm \:  =  \: (3 + 4cot\theta )/(3 - 4cot\theta )`

`\rm \:  =  \: (3 + 4 * (4)/(3) )/(3 - 4 * (4)/(3) )`

`\rm \:  =  \: (3 + (16)/(3) )/(3 - (16)/(3) )`

`\rm \:  =  \: ((9 + 16)/(3) )/( (9 - 16)/(3) )`

`\rm \:  =  \: (25)/( - 7)`

`\rm \:  =  \: - \: (25)/(7)`

Hence,

`\rm \longmapsto\:\boxed{\tt{ (3sin\theta + 4cos\theta )/(3sin\theta - 4cos\theta ) = - (25)/(7) \: }} \\`

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1