1-sin- 1-sin- is equal toa -tan-b -tan-c 2-tan 2-d 2-tan 2-

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Calculating Heights and Distances

1+sinθ 1-sinθ...

Question

$\sqrt{\frac{1 + \sin θ}{1 - \sin θ}}$ is equal to

(a) sec θ + tan θ
(b) sec θ − tan θ
(c) sec² θ + tan² θ
(d) sec² θ − tan² θ

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Solution

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Similar questions

If cot $θ = \frac{7}{8}$ ,evaluate:

(i) $\frac{(1 + s i n θ) (1 - s i n θ)}{(1 + c o s θ) (1 - c o s θ)}$

(ii) $c o t^{2} θ$

\frac{1}{1 - \sin θ} + \frac{1}{1 + \sin θ} = 2 \sec^{2} θ

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

\frac{\tan^{3} θ - 1}{\tan θ - 1} = \sec^{2} θ + \tan θ

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

θ = 178 °

sin θ \sqrt{1 + {cot}^{2} θ} + cos θ \sqrt{1 + {tan}^{2} θ}

\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}

{sec}^{2} θ = 1 + {tan}^{2} θ

{cot}^{2} θ = \frac{7}{8}

0 < θ < 90^{o}

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

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