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3 years ago

2. Here is a picture of an older version of the flag of Great Britain. There is a rigid

Mathematics

1 answer:

Nitella [24]3 years ago

5 0

Answer:

no.

Step-by-step explanation:

the 4 on the sides are half the area of the ones of the top and bottom because if you take two small ones they will fit with one of the big ones.

sorry if I'm wrong.

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Given that cot θ = 1/√5, what is the value of (sec²θ - cosec²θ)/(sec²θ + cosec²θ) ?

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$\mathsf{Given :\;\dfrac{{sec}^2\theta - co{sec}^2\theta}{{sec}^2\theta + co{sec}^2\theta}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}$

$\mathsf{\implies \dfrac{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}{\dfrac{1}{cos^2\theta} + \dfrac{1}{sin^2\theta}}}$

$\mathsf{\implies \dfrac{\dfrac{sin^2\theta - cos^2\theta}{sin^2\theta.cos^2\theta}}{\dfrac{sin^2\theta + cos^2\theta}{sin^2\theta.cos^2\theta}}}$

$\mathsf{\implies \dfrac{sin^2\theta - cos^2\theta}{sin^2\theta + cos^2\theta}}$

Taking sin²θ common in both numerator & denominator, We get :

$\mathsf{\implies \dfrac{sin^2\theta\left(1 - \dfrac{cos^2\theta}{sin^2\theta}\right)}{sin^2\theta\left(1 + \dfrac{cos^2\theta}{sin^2\theta}\right)}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{cot\theta = \dfrac{cos\theta}{sin\theta}}}}$

$\mathsf{\implies \dfrac{1 -cot^2\theta}{1 + cot^2\theta}}$

$\mathsf{Given :\;cot\theta = \dfrac{1}{\sqrt{5}}}$

$\mathsf{\implies \dfrac{1 - \left(\dfrac{1}{\sqrt{5}}\right)^2}{1 + \left(\dfrac{1}{\sqrt{5}}\right)^2}}$

$\mathsf{\implies \dfrac{1 - \dfrac{1}{5}}{1 + \dfrac{1}{5}}}$

$\mathsf{\implies \dfrac{\dfrac{5 - 1}{5}}{\dfrac{5 + 1}{5}}}$

$\mathsf{\implies \dfrac{5 - 1}{5 + 1}}$

$\mathsf{\implies \dfrac{4}{6}}$

$\mathsf{\implies \dfrac{2}{3}}$

Hence, option (a) 2/3 is your correct answer.

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