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Ira Lisetskai [31]
3 years ago
6

-4(5+(-2)) what is the answer

Mathematics
1 answer:
avanturin [10]3 years ago
3 0

Answer:

Step-by-step explanation:

-12

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Prove that<br>{(tanθ+sinθ)^2-(tanθ-sinθ)^2}^2 =16(tanθ+sinθ)(tanθ-sinθ)
USPshnik [31]

First, expand the terms inside the bracket you will get

(( \tan {}^{2} (x)  + 2 \tan(x)  \sin(x)  +  \sin {}^{2} (x)  - ( \tan {}^{2} (x)  - 2 \tan(x)  +  \sin {}^{2} (x) ) {}^{2}  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

( 4 \tan(x)  \sin(x) ) {}^{2}  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x)  \sin {}^{2} (x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16 \tan {}^{2} (x) (1 -  \cos {}^{2} (x) ) = 16 (\tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan {}^{2} (x)  -   \frac{  \sin {}^{2} (x) \cos {}^{2} ( {x}^{} )  }{ \cos {}^{2} (x) }

16( \tan {}^{2} (x)  -  \sin {}^{2} (x) ) = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x)  = 16( \tan(x)  +  \sin(x) )( \tan(x)  -  \sin(x) )

5 0
2 years ago
I NEED THIS DONE IN 3 HOURS AND I CANT FIGURE IT OUT PLEASE HELP
MAVERICK [17]

1. option 1

2. option 1

3. option 4

4. option 1

5. option c

explanation is in picture attached.

excuse me.for my bad handwriting

8 0
3 years ago
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