HELP! NOW! Prove the trig identity: 3sinx/1-cosx=3cscx+3cotx

Mathematics

1 answer:

vichka [17]3 years ago

5 0

Step-by-step explanation:

Given

To Prove

$\dfrac{3\sin x}{1-\cos x}=3\csc x+3\cot x$

Take the Left side of equation and rationalize

$\Rightarrow \dfrac{3\sin x}{1-\cos x}\times \dfrac{1+\cos x}{1+\cos x}\\\\\Rightarrow \dfrac{3\sin (1+\cos x)}{1-\cos^2 x}\\\\\Rightarrow \dfrac{3\sin (1+\cos x)}{\sin^2 x}\\\\\Rightarrow \dfrac{3\sin x}{\sin ^2x}+\dfrac{3\sin x\cos x}{\sin^2 x}\\\\\Rightarrow \dfrac{3}{\sin x}+\dfrac{3\cos x}{\sin x}\\\\\Rightarrow 3\csc x+3\cot x$

Left side is equal to right side

Hence, proved

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Solve – 4(x + 6) – 6 = 4(x – 1) for x

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4Step-by-step explanation:

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Steve made 9 3/6 cups of pancake batter on a weekend camping trip. He used 3 4/6 cios of batter for breakfast on Saturday. Write

LekaFEV [45]

Answer:

9 3/6 ⇒ 19/2

3 4/6 ⇒ 11/3

Step-by-step explanation:

A fraction greater than one is an improper fraction.

$9\frac{3}{6}$ when converted into an improper fraction is

$9\frac{3}{6} =9+\frac{3}{6} \\\\=\frac{9*6}{6}+ \frac{3}{6}\\\\ =\frac{54}{6}+ \frac{3}{6}=\frac{57}{6}\\\\ =\boxed{\frac{19}{2}. }$

$3\frac{4}{6}$ when converted to a fraction is

$3\frac{4}{6}=3+\frac{4}{6}\\\\=\frac{3*6}{6}+ \frac{4}{\\\\6}\\\\=\frac{18}{6}+\frac{4}{6}=\frac{22}{6}\\\\=\boxed{\frac{11}{3}. }$

Thus fractions $9\frac{3}{6}$ and $3\frac{4}{6}$ when converted to fractions greater than one are $\frac{19}{2}$ and $\frac{11}{3}$ respectively.