Math, asked by rahul09652, 1 year ago

Prove that............​

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Answered by Anonymous
5

Answer:

Multiply numerator and denominator by (cosec θ + cot θ).

Then the numerator is (cosec θ + cot θ)², which is what we want to finish with, so the denominator had better be equal to 1.

The denomiator is then

(cosec θ + cot θ)(cosec θ - cot θ)

= cosec² θ - cot² θ

= 1 / sin² θ  -  cos² θ / sin² θ

= ( 1 - cos² θ ) / sin² θ

= sin² θ / sin² θ

= 1.

That finishes the proof of the first equation.

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Next

(cosec θ + cot θ)²

= cosec² θ + cot² θ + 2 cosec θ cot θ

= 1 / sin² θ + cot² θ + 2 cosec θ cot θ

= ( sin² θ + cos² θ ) / sin² θ + cot² θ + 2 cosec θ cot θ

= 1 + cot² θ + cot² θ + 2 cosec θ cot θ

= 1 + 2 cot² θ + 2 cosec θ cot θ

Hope that helps.  Please mark it Brainliest!!!!

Answered by sanjeevravish321
0

Answer:mark me mark me mark me

Step-by-step explanation:

Multiply numerator and denominator by (cosec θ + cot θ).

Then the numerator is (cosec θ + cot θ)², which is what we want to finish with, so the denominator had better be equal to 1.

The denomiator is then

(cosec θ + cot θ)(cosec θ - cot θ)

= cosec² θ - cot² θ

= 1 / sin² θ  -  cos² θ / sin² θ

= ( 1 - cos² θ ) / sin² θ

= sin² θ / sin² θ

= 1.

That finishes the proof of the first equation.

----------------

Next

(cosec θ + cot θ)²

= cosec² θ + cot² θ + 2 cosec θ cot θ

= 1 / sin² θ + cot² θ + 2 cosec θ cot θ

= ( sin² θ + cos² θ ) / sin² θ + cot² θ + 2 cosec θ cot θ

= 1 + cot² θ + cot² θ + 2 cosec θ cot θ

= 1 + 2 cot² θ + 2 cosec θ cot θ

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