Question

prove that 1+cot2 theta/1+cosec2theta = cosec theta​

Answer 1

Correct Question:

To Prove :

${\sf{ 1 + {\dfrac{cot^2 \theta}{1 + cosec \theta}} = cosec \theta}}$

Step-by-step explanation:

L.H.S. = ${\sf{\ \ 1 + {\dfrac{cot^2 \theta}{1 + cosec \theta}}}}$

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${\boxed{\tt{\bigstar \ \ Identity \ : \ 1 + cot^2 \theta = cosec^2 \theta}}}$

${\tt{From \ this, \ we \ get \ [ cot^2 \theta = cosec^2 \theta - 1 ] }}$

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$\implies{\sf{ 1 + {\dfrac{cosec^2 \theta - 1}{1 + cosec \theta}}}}$

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We can write this as :

$\implies{\sf{ 1 + {\dfrac{(cosec \theta)^2 - (1)^2}{1 + cosec \theta}}}}$

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${\boxed{\tt{\bigstar \ \ Identity \ : \ a^2 - b^2 = (a - b)(a + b)}}}$

${\tt{\quad Here, \ a = cosec \theta , \ b = 1}}$

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$\implies{\sf{ 1 + {\dfrac{(cosec \theta - 1)(cosec \theta + 1)}{1 + cosec \theta}}}}$

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Cancelling out the common terms.

$\implies{\sf{ 1 + (cosec \theta - 1)}}$

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Opening the bracket.

$\implies{\sf{ 1 + cosec \theta - 1}}$

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$\implies{\sf{cosec \theta}}$

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= R.H.S.

Hence, verified !!

Answer 2

$\bf To\ prove:\ \tt 1\ +\ \dfrac{cot^2\ \theta}{1\ +\ cosec\ \theta}\ =\ cosec\ \theta.\\\\\\\bf Left\ hand\ side:\\\\\\\tt 1\ +\ \dfrac{cot^2\ \theta}{1\ +\ cosec\ \theta}\\\\\\\\\\\sf We\ know\ that\ cosec^2\ \theta\ -\ cot^2\ \theta =\ 1.\\\\Therefore,\ cosec^2\ \theta\ -\ 1\ =\ cot^2\ \theta \\\\\\\\1\ +\ \dfrac{cosec^2\ \theta\ -\ 1}{1\ +\ cosec\ \theta}\\\\\\\\\ \implies\ 1\ +\ \ \dfrac{(cosec\ \theta)^2\ -\ (1)^2}{1\ +\ cosec\ \theta}\ \ \ \ \ \ \ \bigg[(a)^2\ -\ (b)^2\ =\ (a\ +\ b)(a\ - \ b)\bigg]$

$\tt\Longrightarrow 1\ +\ \dfrac{(cosec\ \theta\ +\ 1)(cosec\ \theta\ -\ 1)}{1\ +\ cosec\ \theta}\\\\\\\\=\ \ \ \ 1\ +\ cosec\ \theta\ -\ 1\\\\\\\\=\ \ \ \ \tt cosec\ \theta\\\\\\\\= \bf Right\ hand\ side.$