Question

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Proof the given identity.

(cosecx - cotx)² = (1 - cosx)/(1 + cosx)

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Answer 1

Multiply by 1-cosX in both numerator and denominator.

{(1-cosx) × (1-cosx) }/{(1+cosx)×(1-cosx)}

Now, you can see in numerator it is (1-cosx)^2

So, expend it as

(a-b)^2=a^2+b^2–2×a×b

And in denominator compress it as

(a-b )(a+b)=a^2-b^2

Now , (1+cos^2x-2×cosx)/(1-cos^2x)

There is another formula we use in denominator to compress it.

Sin^2x+cos^2x=1

1-cos^2x=sin^2x

Now,(1+cos^2x-2×cosx)/sin^2x

Divide each with sin^2x to get the result.

I.e ,1/sin^2x+cos^2x/sin^2x-2×cosx/sin^2x

I.e ,Cosec^2x+cot^2x-2×cotx×cosecx

This is the solution of the given question.

Last line solution formula:

Sinx × cosecx =1

Or, cosecx= 1/sinx

On squaring both side,

Cosec^2x=1/sin^2x

Cosx/sinx= cotx

On squaring both side,

Cos^2x/sin^2x= cot^2x

2×cosx/sinx × 1/sinx

I.e, 2× cotx×cosecx

Thank you.

Answer 2

★ Concept :-

Here the concept of Trigonometric Identities has been used. We see that we are given an equation where we need to prove that L.H.S. is equal to R.H.S. This can be easily done by taking one at a time. Firstly we can take L.H.S. and then we can simplify it by using identities and then apply to trignometric ratios. And finally we can convert it into the form of R.H.S.

Let's do it !!

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★ Solution :-

Given to prove,

$\;\bf{\mapsto\;\;\blue{(cosec\:x\:-\:\cot x)^{2}\;=\;\dfrac{(1\:-\:\cos x)}{(1\:+\:\cos x)}}}$

From here we have,

$\;\sf{\rightarrow\;\;\orange{L.H.S.\;=\;(cosec\:x\:-\:\cot x)^{2}}}$

And,

$\;\sf{\rightarrow\;\;\green{R.H.S.\;=\;\dfrac{(1\:-\:\cos x)}{(1\:+\:\cos x)}}}$

Let's now firstly simplify the L.H.S. part .

Then,

$\;\tt{\rightarrow\;\;L.H.S.\;=\;(cosec\:x\:-\:\cot x)^{2}}$

We know that,

• cosec A = 1/(sin A)

• cot A = (cos A)/(sin A)

Here A = x (for this identity)

By applying this here, we get

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\bigg(\dfrac{1}{\sin x}\:-\:\dfrac{\cos x}{\sin x}\bigg)^{2}}$

We know that,

• (a - b)² = a² + b² - 2ab

Here a = 1/(sin x) and b = (cos x)/(sin x)

By applying this here, we get

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\bigg[\bigg(\dfrac{1}{\sin x}\bigg)^{2}\:+\:\bigg(\dfrac{\cos x}{\sin x}\bigg)^{2}\:-\:2\bigg(\dfrac{1}{\sin x}\bigg)\bigg(\dfrac{\cos x}{\sin x}\bigg)\bigg]}$

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\bigg[\dfrac{1}{\sin^{2}x}\:+\:\dfrac{\cos^{2}x}{\sin^{2}x}\:-\:\dfrac{2\cos x}{\sin^{2}x}\bigg]}$

Now the denominator of all three terms is sin²x . So we can apply algebraic functions.

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{1\:+\:\cos^{2}x\:-\:2\cos x}{\sin^{2}x}}$

We know that,

• a² + b² - 2ab = (a - b)²

Here a = 1 and b = cos x

By applying this, we get

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{(1\:-\:\cos x)^{2}}{\sin^{2}x}}$

We have one identity,

• sin² A = 1 - cos² A

Here A = x

So using this, we get

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{(1\:-\:\cos x)^{2}}{1\:-\:\cos^{2}x}}$

This can be written as,

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{(1\:-\:\cos x)(1\:-\:\cos x)}{(1)^{2}\:-\:(\cos x)^{2}}}$

We know that,

• a² - b² = (a + b)(a - b)

So applying this here, we get

$\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{(1\:-\:\cos x)(1\:-\:\cos x)}{(1\:-\:\cos x)(1\:+\:\cos x)}}$

By cancelling the like terms, we get

$\;\bf{\rightarrow\;\;\red{L.H.S.\;=\;\dfrac{(1\:-\:\cos x)}{(1\:+\:\cos x)}}}$

Clearly L.H.S. = R.H.S.

So,

$\;\bf{\rightarrow\;\;\purple{R.H.S.\;=\;L.H.S.\;=\;\dfrac{(1\:-\:\cos x)}{(1\:+\:\cos x)}}}$

• Hence, proved

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★ More to know :-

• sec² A = 1 - tan² A

• cosec² A = 1 - cot² A

• sin A = cos(90° - A)

• cosec A = sec(90° - A)

• tan A = cot(90° - A)