secθ −1tanθ +secθ +1tanθ is equal to

sec2θ −1tanθ (secθ +1+secθ −1) =sec2θ −1 2tanθ secθ =tan2θ 2tanθ secθ ⇒tanθ 2secθ =2cosθ .cosθ sinθ 1=2cosecθ

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{ { } }{ { } -1 } + { { } }{ { } +1 } is equal

Question

$\frac{tan θ}{sec θ - 1} + \frac{tan θ}{sec θ + 1}$ is equal to

cosec θ

2 cosec θ

sec θ

2 sec θ

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Michael Discussed

\frac{tan θ}{sec θ - 1} + \frac{tan θ}{sec θ + 1}

is equal to

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10 mins ago

\frac{tan θ ( sec θ + 1 + sec θ - 1 )}{sec ^{2} θ - 1} = \frac{2 tan θ sec θ}{sec ^{2} θ - 1} = \frac{2 tan θ sec θ}{tan ^{2} θ}

\Rightarrow \frac{2 sec θ}{tan θ} = 2 \frac{1}{cos θ . \frac{sin θ}{cos θ}} = 2 cosec θ

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Question Text	$\frac{tan θ}{sec θ - 1} + \frac{tan θ}{sec θ + 1}$ is equal to
Answer Type	Text solution:1
Upvotes	150