√(θ 1 ) #160; √(θ +1 ) #160; =θ b tanθ #160; or, θ 1 √(^ 2 θ +1 ) #160; =θ b tanθ #160; #160;θ 1 tanθ #160; =θ b tanθ #160; ...

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Trigonometric Identities

If √θ-1√θ+1...

Question

If $\frac{\sqrt{sec θ - 1}}{\sqrt{sec θ + 1}} = \frac{sec θ - b}{tan θ}$ .Find $b$

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Solution

$\frac{\sqrt{sec θ - 1}}{\sqrt{sec θ + 1}} = \frac{sec θ - b}{tan θ}$
or, $\frac{sec θ - 1}{\sqrt{{sec}^{2} θ + 1}} = \frac{sec θ - b}{tan θ}$
$⟹ \frac{sec θ - 1}{tan θ} = \frac{sec θ - b}{tan θ} = b = 1$

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Similar questions

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

Is the identity $\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ} = 1 + tan θ + cot θ$ true or false?

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sec θ = \sqrt{2}

\frac{3 π}{2} < θ < 2 π

\frac{1 + tan θ + csc θ}{1 + cot θ - csc θ} = m

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