Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$$(cosec\; θ – cot \;θ)^2 =\dfrac{(1-cos\; θ)}{(1+cos\; θ)}$$

Asked by Abhisek | 1 year ago |  89

##### Solution :-

To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)

L.H.S. = (cosec θ – cot θ)2

The above equation is in the form of (a-b)2, and expand it

Since (a-b)2 = a2 + b2 – 2ab

Here a = cosec θ and b = cot θ

= (cosec2θ + cot2θ – 2cosec θ cot θ)

Now, apply the corresponding inverse functions and equivalent ratios to simplify

$$\dfrac{1}{sin^2θ}+ \dfrac{cos^2θ}{sin^2θ}- \dfrac{2cos θ}{sin^2θ}$$

$$\dfrac{(1 + cos^2θ – 2cos θ)}{(1 – cos^2θ)}$$

$$\dfrac{(1-cos \;θ)^2}{(1 – cos\;θ)(1+cos\; θ)}$$

$$\dfrac{ (1-cos\; θ)}{(1+cos\; θ)}$$ = R.H.S.

Therefore, (cosec θ – cot θ)

$$\dfrac{ (1-cos\; θ)}{(1+cos\; θ)}$$

Hence proved.

Answered by Pragya Singh | 1 year ago

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