Question

Trigonometry !…
$\sf{ If \ \pi \leq \theta \leq \dfrac{3 \pi}{2} , \ then }\\ \sf{ \sqrt{ \frac{1 + cos \: \theta}{1 - cos \: \theta} + \sqrt{ \frac{1 - cos \: \theta}{1 + cos \: \theta} } } \: is \: equal \: to \: : } \\ \\ \sf{ a) 2 \: cosec \: \theta} \\ \sf{b) - 2 \: cosec \: \theta}\\ \sf{ c)2 \: sec \: \theta} \\ \sf{d) - \: sec \: \theta}$
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Answer 1

Given :

Correct question $\sf\sqrt{\dfrac{1+cos\theta}{1-cos\theta}}+\sqrt{\dfrac{1-cos\theta}{1+cos\theta}}$

a) 2 cosec θ

b) -2 cosec θ

c) 2 sec θ

d) - sec θ

According to the question :

Solving :

⟹ $\sf\sqrt{\dfrac{1+cos\:\theta}{1-cos\:\theta}}+\sqrt{\dfrac{1-cos\:\theta}{1+cos\:\theta}}$

Now Rationalise the denominator,

⟹ $\sf\sqrt{\dfrac{1+cos\:\theta}{1-cos\:\theta}\times\dfrac{1+cos\:\theta}{1+cos\:\theta}}+\sqrt{\dfrac{1-cos\:\theta}{1+cos\:\theta}\times\dfrac{1-cos\:\theta}{1-cos\:\theta}}$

Then ,

⟹ $\sf\sqrt{\dfrac{(1+cos\:\theta)^2}{1-cos^2\:\theta}}+\sqrt{\dfrac{(1-cos\:\theta)^2}{1-cos^2\:\theta}}</p><p>$

⟹ $\sf\sqrt{\dfrac{(1+cos\:\theta)^2}{sin^2\:\theta}}+\sqrt{\dfrac{(1-cos\:\theta)^2}{sin^2\:\theta}}$

⟹ $\sf\dfrac{1+cos\:\theta}{sin\:\theta}+\dfrac{1-cos\:\theta}{sin\:\theta}$

Now take LCM then ,

⟹ $\sf\dfrac{1+cos\:\theta+1-cos\:\theta}{sin\:\theta}$

⟹ $\sf\dfrac{2}{sin\:\theta}$ $\sf=2\csc\theta$

So It's Done !!

Answer 2

Answer:

answer for the given problem is given