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ivanzaharov [21]
4 years ago
13

Order from least to greatest -2, 5, 1, 0, -3

Mathematics
2 answers:
Vedmedyk [2.9K]4 years ago
7 0

Answer:-3,-2,0,1,5

Step-by-step explanation:

VLD [36.1K]4 years ago
5 0
-3,-2,0,1,5 is the answer

i know this because i’m smart
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Prove that (sec theta-cos theta) ( cot theta+tan theta) = tan theta ×sec theta​
meriva

Given:

(\sec \theta \cos \theta)(\cot \theta+\tan \theta)=\tan \theta \sec \theta

To prove:

The given statement.

Proof:

We have,

(\sec \theta -\cos \theta)(\cot \theta+\tan \theta)=\tan \theta \sec \theta

Taking LHS, we get

LHS=(\sec \theta -\cos \theta)(\cot \theta+\tan \theta)

LHS=(\dfrac{1}{\cos \theta }-\cos \theta)(\dfrac{\cos \theta}{\sin \theta}+\dfrac{\sin \theta}{\cos \theta})

LHS=(\dfrac{1-\cos^2 \theta }{\cos \theta })(\dfrac{\cos^2 \theta+\sin^2 \theta}{\sin \theta\cos \theta})

LHS=(\dfrac{\sin^2 \theta }{\cos \theta })(\dfrac{1}{\sin \theta\cos \theta})          [\because \cos^2 \theta+\sin^2 \theta=1]

LHS=(\dfrac{\sin \theta }{\cos \theta })(\dfrac{1}{\cos \theta})

LHS=\tan \theta \sec \theta

LHS=RHS

Hence proved.

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3 years ago
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