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Prove that: 2 + ^ { 2 } + { 1 } { ^ { 2 } } = { 1 } {

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Prove that:
$2 + tan^{2} θ + \frac{1}{tan ^{2} θ} = \frac{1}{sin ^{2} θ - sin ^{4} θ}$

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Prove that:

2 + tan^{2} θ + \frac{1}{tan ^{2} θ} = \frac{1}{sin ^{2} θ - sin ^{4} θ}

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Text solutionVerified

2 + t a n^{2} θ + \frac{1}{t a n ^{2} θ}

\Rightarrow 2 + \frac{sin ^{2} θ}{cos ^{2} θ} + \frac{1}{\frac{sin ^{2} θ}{cos ^{2} θ}}

\Rightarrow \frac{2 ( cos ^{2} θ ) + sin ^{2} θ}{cos ^{2} θ} + \frac{cos ^{2} θ}{sin ^{2} θ}

\Rightarrow \frac{2 ( 1 - sin ^{2} θ ) + sin ^{2} θ}{( 1 - sin ^{2} θ )} + \frac{( 1 - sin ^{2} θ )}{sin ^{2} θ}

\Rightarrow \frac{2 - 2 sin ^{2} θ + sin ^{2} θ}{1 - sin ^{2} θ} + \frac{1 - sin ^{2} θ}{sin ^{2} θ}

\Rightarrow \frac{2 sin ^{2} θ - sin ^{4} θ + 1 + sin ^{4} θ - 2 sin ^{2} θ}{sin ^{2} θ - sin ^{4} θ}

\Rightarrow \frac{1}{sin ^{2} θ - sin ^{4} θ}

Hence proved.

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Question Text	Prove that: $2 + tan^{2} θ + \frac{1}{tan ^{2} θ} = \frac{1}{sin ^{2} θ - sin ^{4} θ}$
Answer Type	Text solution:1
Upvotes	150