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Over [174]
3 years ago
15

If tanA=a then find sin4A-2sin2A/ sin4A+2sin2A​

Mathematics
1 answer:
anygoal [31]3 years ago
6 0

Answer:

The value of the given expression is

\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

Step by step Explanation:

Given that tanA=a

To find the value of \frac{sin4A-2sin2A}{sin4A+2sin2A}

Let us find the value of the expression :

\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A} ( by using the formula sin2A=2cosAsinA here A=2A)  

=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}

=\frac{(cos2A-1)}{(cos2A+1)}

=\frac{(-(1-cos2A))}{(1+cos2A)}(using  sin^2A+cos^2A=1  here A=2A)

=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}(using cos2A=cos^2A-sin^2A here A=2A)

=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}

=\frac{-(sin^2A+sin^2A)}{cos^2A+cos^2A}

=\frac{-2sin^2A}{2cos^2A}

=-\frac{sin^2A}{cos^2A}

=-tan^2A  ( using tanA=\frac{sinA}{cosA} here A=2A )

 =-a^2 (since tanA=a given )

Therefore \frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2

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2 years ago
What are the zeros of the function?
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The function is:

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2 years ago
I will give a brainlest, i promise, please help me with this.​
valkas [14]

Answer:

1) b and m

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Step-by-step explanation:

1) all straight lines sum to 180°

subtract the angles given from 180°

the other angle for b is 25°, while the other angle for m is 155°

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2) ∠8 should be the same as ∠6, ∠10 should be the same as ∠3, ∠7 same as ∠5 and ∠9 same as ∠4

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we get x=20(nearest whole number)

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4) since we need to show that they are parallel,

(2x+30)°=(4x-90)°

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x=60

we then plug the x value into the two equations, in which we get 150° for both the angles [2(60)+30=4(60)-90] ⇒ (150=150)

I hope u understand it the way I put it.

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Answer:

See photo

Step-by-step explanation:

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