Given that sin\(\theta\)=a/b ,then tan\(\theta\) is equal to(a) \(\c

1.	Given that sin\(\theta\)=a/b ,then tan\(\theta\) is equal to(a) \(\cfrac{b}{\sqrt{a^2+b^2}}\)(b) \(\cfrac{b}{\sqrt{b^2-a^2}}\)(c) \(\cfrac{a}{\sqrt{a^2-b^2}}\)(d) \(\cfrac{a}{\sqrt{b^2-a^2}}\)
Answer» Correct answer is: (d) \(\cfrac{a}{\sqrt{b^2-a^2}}\) sin\(\theta\) = a/b H²=P²+B² b²=a²+B² B=\(\sqrt{b^2 -a^2}\) tan\(\theta\) = P/B=a/\(\sqrt{b^2 -a^2}\)

Given that sin\(\theta\)=a/b ,then tan\(\theta\) is equal to(a) \(\cfrac{b}{\sqrt{a^2+b^2}}\)(b) \(\cfrac{b}{\sqrt{b^2-a^2}}\)(c) \(\cfrac{a}{\sqrt{a^2-b^2}}\)(d) \(\cfrac{a}{\sqrt{b^2-a^2}}\)

Answer»

Correct answer is: (d) \(\cfrac{a}{\sqrt{b^2-a^2}}\)
sin\(\theta\) = a/b

H²=P²+B²

b²=a²+B²

B=\(\sqrt{b^2 -a^2}\)

tan\(\theta\) = P/B=a/\(\sqrt{b^2 -a^2}\)

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Given that sin\(\theta\)=a/b ,then tan\(\theta\) is equal to(a) \(\cfrac{b}{\sqrt{a^2+b^2}}\)(b) \(\cfrac{b}{\sqrt{b^2-a^2}}\)(c) \(\cfrac{a}{\sqrt{a^2-b^2}}\)(d) \(\cfrac{a}{\sqrt{b^2-a^2}}\)

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