Question

If `sin θ = 6/10`, find without using table, the value of (cos θ + tan θ).

Answer 1

Given: `sin θ = 6/10` = `3/5`.

Step-wise calculation:

1. `cos θ = ± sqrt(1 - sin^2θ)` 

= `± sqrt(1 - (3/5)^2)` 

= `± sqrt(1 - 9/25)` 

= `± sqrt(16/25)` 

= `± 4/5`

2. `tan θ = (sin θ)/(cos θ)`   ...(Same sign as cos θ)

= `(3/5)/(± 4/5)`

= `± 3/4`

3. `cos θ + tan θ = cos θ + (sin θ/cos θ)`

= `(cos^2θ + sin θ)/(cos θ)`

cos²θ = 1 – sin²θ

= `16/25`

So, `cos^2θ + sin θ = 16/25 + 3/5`

= `16/25 + 15/25`

= `31/25`

Therefore, `cos θ + tan θ = (31/25)/(±4/5) = ± 31/20`.

If θ is acute (cos θ > 0): cos θ + tan θ = `31/20`.

If θ is in the quadrant where cos θ < 0: cos θ + tan θ = `-31/20`.