- -78- then evaluate 1-sin-1-sin-1-cos-1-cos-

θ =78=Baseperpendicular ⇒ Hypotenuse =√(p^2+B^2)=√(7^2+8^2)=√(49+64) =√(113) =√(113) ∴(1+sinθ)(1 sinθ)(1+cosθ)(1 cosθ) =1 s ...

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General Solution of Trigonometric Equation

θ =78, then e...

Question

$cot θ = \frac{7}{8}$ , then evaluate $\frac{(1 + sin θ) (1 - sin θ)}{(1 + cos θ) (1 - cos θ)}$ .

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If $cot θ = \frac{7}{8}$ , evaluate: $\frac{(1 + sin θ) (1 - sin θ)}{(1 + cos θ) (1 - cos θ)}$

cot θ = \frac{7}{8}

\frac{(1 + cos θ) (1 - cos θ)}{(1 - sin θ) (1 + sin θ)}

\cot θ = \frac{15}{8} then evaluate \frac{(1 + \sin θ) (1 - \sin θ)}{(1 + \cos θ) (1 - \cos 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}

(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

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