Question

If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ.

Answer 1

5 sec θ – 12 cosec θ = 0   ...[Given]

∴ 5 sec θ = 12 cosec θ

∴ `5/(cosθ) = 12/(sinθ)`   ...`[∵ sec θ = 1/(cosθ), "cosec"  θ = 1/(sin θ)]`

∴ `(sinθ)/(cosθ) = 12/5`

∴ `tan θ = 12/5`

We know that,

1 + tan²θ = sec²θ

∴ `1 + (12/5)^2 = sec^2θ`

∴ `1 + 144/25 = sec^2θ`

∴ `(25 + 144)/25 = sec^2θ`

∴ `sec^2θ = 169/25`

∴ `sec θ = 13/5`   ...[Taking square root of both sides]

Now, `cos θ = 1/(sec θ)`

= `1/((13/5))`

∴ `cos θ = 5/13`

We know that,

sin²θ + cos²θ = 1

∴ `sin^2θ + (5/13)^2 = 1`

∴ `sin^2θ + 25/169 = 1`

∴ `sec^2θ = 1 - 25/169`

∴ `sec^2θ = (169 - 25)/169`

∴ `sec^2θ = 144/169`

∴ `sin θ = 12/13`   ...[Taking square root of both sides]

∴ `sin θ = 12/13, sec θ = 13/5`