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प्रश्न
Given sec θ = `13/12`, calculate all other trigonometric ratios.
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उत्तर
Let ΔABC be a right-angled triangle, right angled at point B.
It is given that:
sec θ = `"hypotenuse"/"side adjacent to ∠θ" = "AC"/"AB" = 13/12`
Let AC = 13k and AB = 12k, where k is a positive integer.
Applying pythagoras theorem in Δ ABC, we obtain:
AC2 = AB2 + BC2
BC2 = AC2 - AB2
BC2 = (13k)2 - (12k)2
BC2 = 169 k2 - 144 k2
BC2 = 25k2
BC = 5k
sin θ = `("side opposite to ∠θ")/("hypotenuse") = ("BC")/("AC") = 5/13`
cos θ = `("side adjacent to ∠θ")/("hypotenuse") = ("AB")/("AC") = 12/13`
tan θ = `("side opposite to ∠θ")/("side adjacent to ∠θ") = "(BC)"/"(AB)" = 5/12`
cot θ = `("side adjacent to ∠θ")/("side opposite to ∠θ") = ("AB")/("BC") = 12/5`
cosec θ = `("hypotenuse")/("side opposite to ∠θ") = ("AC")/("BC") = 13/5`
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