Question

if tan theta=1÷✓5 what is the value of
cosec square theta-secsquaretheta÷cosec square theta+sec square theta​

Answer 1

Question:

If $\sf\:\tan\:\theta\:=\:\dfrac{1}{\sqrt{5}}$, find the value of

$\displaystyle\sf\:\dfrac{\csc^2\:\theta\:-\:\sec^2\:\theta}{\csc^2\:\theta\:+\:\sec^2\:\theta}$

Answer:

$\displaystyle\boxed{\red{\sf\:\dfrac{\csc^2\:\theta\:-\:\sec^2\:\theta}{\csc^2\:\theta\:+\:\sec^2\:\theta}\:=\:\dfrac{2}{3}}}$

Step-by-step-explanation:

We have given that,

$\sf\:\tan\:\theta\:=\:\dfrac{1}{\sqrt{5}}$

We know that,

$\displaystyle\pink{\sf\:\cot\:\theta\:=\:\dfrac{1}{\tan\:\theta}}\sf\:\:\:-\:-\:[\:Trigonometric\:identity\:]$

$\displaystyle\implies\sf\:\cot\:\theta\:=\:\dfrac{1}{\dfrac{1}{\sqrt{5}}}$

$\displaystyle\implies\sf\:\cot\:\theta\:=\:\dfrac{1}{1}\:\times\:\dfrac{\sqrt{5}}{1}$

$\displaystyle\implies\sf\:\cot\:\theta\:=\:\dfrac{\sqrt{5}}{1}$

$\displaystyle\implies\boxed{\red{\sf\:\cot\:\theta\:=\:\sqrt{5}}}$

Now, we know that,

$\displaystyle\pink{\sf\:\csc^2\:\theta\:=\:1\:+\:\cot^2\:\theta}\sf\:\:\:-\:-\:[\:Trigonometric\:identity\:]$

$\displaystyle\implies\sf\:\csc^2\:\theta\:=\:1\:+\:(\:\sqrt{5}\:)^2$

$\displaystyle\implies\sf\:\csc^2\:\theta\:=\:1\:+\:5$

$\displaystyle\implies\boxed{\red{\sf\:\csc^2\:\theta\:=\:6}}$

Now, we know that,

$\displaystyle\pink{\sf\:\sec^2\:\theta\:=\:1\:+\:\tan^2\:\theta}\sf\:\:\:-\:-\:-\:[\:Trigonometric\:identity\:]$

$\displaystyle\implies\sf\:\sec^2\:\theta\:=\:1\:+\:\left(\:\dfrac{1}{\sqrt{5}}\:\right)^2$

$\displaystyle{\implies\sf\:\sec^2\:\theta\:=\:1\:+\:\left(\:\dfrac{\:(\:1\:)^2}{\;(\:\sqrt{5}\:)^2}\:\right)\:\:\:-\:-\:[\:\because\:\left(\:\dfrac{a}{b}\:\right)^2\:=\:\dfrac{a^2}{b^2}\:]}$

$\displaystyle\implies\sf\:\sec^2\:\theta\:=\:1\:+\:\dfrac{1}{5}$

$\displaystyle\implies\sf\:\sec^2\:\theta\:=\:\dfrac{1\:\times\:5\:+\:1}{5}$

$\displaystyle\implies\sf\:\sec^2\:\theta\:=\:\dfrac{5\:+\:1}{5}$

$\displaystyle\implies\boxed{\red{\sf\:\sec^2\:\theta\:=\:\dfrac{6}{5}}}$

Now,

$\displaystyle\sf\:\dfrac{\csc^2\:\theta\:-\:\sec^2\:\theta}{\csc^2\:\theta\:+\:\sec^2\:\theta}$

$\displaystyle\implies\sf\:\dfrac{6\:-\:\dfrac{6}{5}}{6\:+\:\dfrac{6}{5}}$

$\displaystyle\implies\sf\:\dfrac{\dfrac{6\:\times\:5\:-\:6}{5}}{\dfrac{6\:\times\:5\:+\:6}{5}}$

$\displaystyle\implies\sf\:\dfrac{\dfrac{30\:-\:6}{5}}{\dfrac{30\:+\:6}{5}}$

$\displaystyle\implies\sf\:\dfrac{\dfrac{24}{5}}{\dfrac{36}{5}}$

$\displaystyle\implies\sf\:\dfrac{24}{\cancel{5}}\:\times\:\dfrac{\cancel{5}}{36}$

$\displaystyle\implies\sf\:\cancel{\dfrac{24}{36}}$

$\displaystyle\therefore\boxed{\red{\sf\:\dfrac{\csc^2\:\theta\:-\:\sec^2\:\theta}{\csc^2\:\theta\:+\:\sec^2\:\theta}\:=\:\dfrac{2}{3}}}$

Answer 2

Step-by-step explanation:

Given : -

• tan theta=1÷✓5

To find : -

• The value of (cosec²θ - sec²θ)/(cosec²θ + sec²θ)

solution : -

• here tanθ = 1/√5 = p/b

• so, p = 1 and b = √5

Pythagoras theorem, h = √(1² + √5²) = √6

now cosecθ = h/p = √6/1 = √6

secθ = h/b = √6/√5

now (cosec²θ - sec²θ)/(cosec²θ + sec²θ)

= {(√6)² - (√6/√5)²}/{(√6)² + (√6/√5)²}

= (6 - 6/5)/(6 + 6/5)

= (30 - 6)/(30 + 6)

= 24/36

= 2/3

Therefore the value of (cosec²θ - sec²θ)/(cosec²θ + sec²θ) is 2/3