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The correct option is B False 1θ 1 1θ+1 =θ+1 θ+1^2θ 1 =2tan^2θ =2^2θ ...

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Basic Inverse Trigonometric Functions

1 θ - 1 - 1 θ...

Question

$\frac{1}{sec θ - 1} - \frac{1}{sec θ + 1} = cot θ$ .

True

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False

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Solution

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\sqrt{\frac{secθ - 1}{secθ + 1}} + \sqrt{\frac{secθ + 1}{secθ - 1}} = ?

\sqrt{\frac{\sec θ - 1}{\sec θ + 1}} + \sqrt{\frac{\sec θ + 1}{\sec θ - 1}} = 2 cosec θ

\sqrt{\frac{1 + \sin θ}{1 - \sin θ}} + \sqrt{\frac{1 - \sin θ}{1 + \sin θ}} = 2 \sec θ

\sqrt{\frac{1 + \cos θ}{1 - \cos θ}} + \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = 2 cosec θ

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

\frac{\sin θ + 1 - \cos θ}{\cos θ - 1 + \sin θ} = \frac{1 + \sin θ}{\cos θ}

\frac{sec θ - 1}{sec θ + 1} = \frac{1 - cos θ}{1 + cos θ}

(sec θ + tan θ - 1) (sec θ - tan θ + 1) = 2 tan θ

\frac{secθ - 1}{secθ + 1} = \frac{\sin^{2} θ}{(1 + cosθ)^{2}}

\frac{secθ - tanθ}{secθ + tanθ} = \frac{\cos^{2} θ}{(1 + sinθ)^{2}}

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