Given - - 7 8- then evaluate i 1-sin- 1-sin- 1-cos- 1-cos- ii 1-sin- cos-

#10;(i) #10; (1 + sinθ )(1 sinθ )(1 + cosθ )(1 #10;cosθ ) #10; #10; =1 sin^2θ1 cos^2θ #10; #10; =cos^2θsin^2θ=cot^2θ=(78)^2=496 ...

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Circular Measurement of Angle

Given θ = 7...

Question

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

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Solution

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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}

(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

c o t θ = \frac{7}{8}

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

s i n θ (1 + s i n θ) + c o s θ (1 + c o s θ) = x

s i n θ (1 - s i n θ) + c o s θ (1 - c o s θ) = y

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