If sin theta=p/q find the value of all trignometric ratios
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6
sin¢=p/q
cosec¢=q/p
cos¢=(√q^2-p^2)/q
sec¢=q/(√q^2-p^2)
tan¢=p/(√q^2-p^2)
cot¢=(√q^2-p^2)/p
cosec¢=q/p
cos¢=(√q^2-p^2)/q
sec¢=q/(√q^2-p^2)
tan¢=p/(√q^2-p^2)
cot¢=(√q^2-p^2)/p
Answered by
4
sin∅ = opposite side/hypotenuse
Here, sin∅=p/q
Therefore, opposite side=p
hypotenuse=q
So, adjacent side=√(q²-p²)
cos∅=√(q²-p²)/q
tan∅ = p/√(q²-p²)
cot∅ = √(q²-p²)/p
sec∅ = q/√(q²-p²)
cosec∅ = q/p
Here, sin∅=p/q
Therefore, opposite side=p
hypotenuse=q
So, adjacent side=√(q²-p²)
cos∅=√(q²-p²)/q
tan∅ = p/√(q²-p²)
cot∅ = √(q²-p²)/p
sec∅ = q/√(q²-p²)
cosec∅ = q/p
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