Sin- 1-cos- 1-tan- is equal toa 0b 1c sin-cos-d sin-cos-

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Range of Trigonometric Ratios from 0 to 90 Degrees

sinθ 1-θ+cosθ...

Question

$\frac{\sin θ}{1 - \cot θ} + \frac{\cos θ}{1 - \tan θ}$ is equal to

(a) 0
(b) 1
(c) sin θ + cos θ
(d) sin θ − cos θ

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Solution

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Similar questions

(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}

cot θ = \frac{7}{8}

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

The expression $E = \frac{sec θ + tan θ - 1}{tan θ - sec θ + 1}$ can be simplified to -

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

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