If - - 7 8 then find the value of 1-sin- 1-sin- 1-cos- 1-cos-

(1+ sin θ )(1 sin θ )(1+ cos θ )(1 cosθ )=1 sin^2θ1 cos^2θ=cos^2θsin^2θ = cot^2θ = (78)^2 =4964 ...

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Trigonometric Identities

If θ = 7 8 ...

Question

If $cot θ = \frac{7}{8}$ then find the value of $\frac{(1 + sin θ) (1 - sin θ)}{(1 + cos θ) (1 - cos θ)}$

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

\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}

If $θ$ is an acute angle such that ${tan}^{2} θ = \frac{8}{7}$ , then the value of $\frac{(1 + sin θ) (1 - sin θ)}{(1 + cos θ) (1 - cos θ)}$ is

A. $\frac{7}{4}$

B. $\frac{7}{8}$

C. $\frac{8}{7}$

D. $\frac{64}{49}$

(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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