Prove that sin-1-tan-cos-1-co-

L:H:S = sinθ (1+tanθ )+cosθ (1+cotθ ) = sinθ (1+sinθcosθ)+cosθ (1+cosθsinθ) = sinθ(sinθ +cosθ )cosθ+cosθ(sinθ +cosθ )sinθ = (sinθ +cosθ )(si ...

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Trigonometric Ratios of Compound Angles

Prove that ...

Question

Prove that $sin θ (1 + tan θ) + cos θ (1 + cot θ) = sec θ + c o sec θ$

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sin θ (1 + tan θ) + cos θ (1 + cot θ) = sec θ + csc θ

(i) $s e c θ (1 - s i n θ) (s e c θ + t a n θ) = 1$
(ii) $s i n θ (1 + t a n θ) + c o s θ (1 + c o t θ) = (s e c θ + c o s e c θ)$

(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

\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}

(1 + tan θ + sec θ) (1 + cot θ - c o sec θ)

\frac{tan A - sin A}{tan A + sin A} = \frac{sec A - 1}{sec A + 1}

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