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Maslowich
3 years ago
14

{p} + p = 4 \\ " alt=" \frac{1}{p} + p = 4 \\ " align="absmiddle" class="latex-formula">
so, now find the value of,
\frac{1}{ {p}^{2} }  +  {p}^{2} =
need the step by step explanation...will give the brainliest!​

Mathematics
1 answer:
Setler [38]3 years ago
6 0

Answer:

The answer is is 14.

Step-by-step explanation:

Firstly, you have to square on both side by using the formula, (a+b)² = a²+2ab+b² :

(1/p + p)² = 4²

1/p² + 2(1/p)(p) + p² = 16

1/p² + 2 + p²= 16

It is given to us to find the value of 1/p² + p² = ? . By doing so, we can eliminate 2 by substracting 2 on both side :

1/p² + 2 + p² - 2 = 16 - 2

1/p² + p = 14

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Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta

To start, we can multiply the fraction by (1 - sin(θ)). This yields:

\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta

Simplify. The denominator uses the difference of two squares pattern:

\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta

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\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta

Split into two separate fractions:

\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta

Rewrite the two fractions:

\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta

By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:

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