Show that - - - 1 - - 1 - 1 - cos- 1 - cos-

secΘ =1/cosΘ 1/cosΘ 1/1/cosΘ+1 =1 cosΘ/cosΘ/1+cosΘ/cosΘ = 1 cosΘ/cosΘ 1 cosΘ/1+cosΘ H.P. ...

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Integration of Trigonometric Functions

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Question

Show that : $\frac{sec θ - 1}{sec θ + 1} = \frac{1 - cos θ}{1 + cos θ}$

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\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{\sin θ - \cos θ + 1}{\sin θ + \cos θ - 1} = \frac{1}{s e c θ - \tan θ}

(i) \sqrt{\frac{1 + \sin θ}{1 - \sin θ}} = (\sec θ + \tan θ) (ii) \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = (cosec θ - \cot θ) (iii) \sqrt{\frac{1 + \cos θ}{1 - \cos θ}} + \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = 2 cosec θ

(1 + {cot}^{2} θ) (1 - cos θ) (1 + cos θ) = 1

\frac{sinθ - cosθ + 1}{sinθ + cosθ - 1} = \frac{1}{(secθ - tanθ)} .

\frac{secθ cosec (90 ° - θ) - \tan θ \cot (90 ° - θ) + \sin^{2} 65 ° + \sin^{2} 25 °}{\tan 10 ° \tan 20 ° \tan 60 ° \tan 70 ° \tan 80 °}

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