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romanna [79]
3 years ago
13

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

Mathematics
1 answer:
Bogdan [553]3 years ago
3 0

Step-by-step explanation:

\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}}

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

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