Find the simplest form of 1-sin- 1-sin- 1-tan 2 -

Formulae: (a+b)(a b)=a^2 b^2 1 sin^2θ=cos^2θ 1+tan^2A=^2A cos A=1 A (1 sinθ)(1+sinθ)(1+tan^2θ)=(1 sin^2θ)(1+tan^2θ) #160; #160; ...

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Basic Inverse Trigonometric Functions

Find the simp...

Question

Find the simplest form of $(1 - sin θ) (1 + sin θ) (1 + {tan}^{2} θ)$

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(1 + {tan}^{2} θ) (1 - sin θ) (1 + sin θ) = 1

\cot θ = \frac{15}{8} then evaluate \frac{(1 + \sin θ) (1 - \sin θ)}{(1 + \cos θ) (1 - \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. $\frac{64}{49}$

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

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