Answer :

We have, cosθ = (7k)/(25k) = base/hypotenuse (For some value of k)

By Pythagoras theorem, (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2}

∴AB^{2} = BC^{2} + AC^{2}

= (25k)^{2} = BC^{2} + (7k)^{2} (for some value of k)

= 625k^{2} = BC^{2} + 49k^{2}

= BC^{2} = 576k^{2}

= BC^{2} = (24k)^{2}

→ BC = 24k

Hence, the trigonometric ratios of the given θ are:

sinθ = BC/AB = (24k)/(25k) = 24/25

cosθ = 7/25

tanθ = BC/AC = sinθ /cosθ = 24/7

cotθ = AC/BC = 1/tanθ = 7/24

cosecθ = AB/BC = 1/sinθ = 25/24

secθ = AB/AC = 1/cosθ = 25/7

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