If - is an acute angle such that tan 2 -87- then the value of 1-sin- 1-sin- 1-cos- 1 -cos - isa 78b 87c 74d 6449

Given that: andis an acute angle We have to find the following expression Since nbsp; Since We know thatand We find: Hence the correct opti ...

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Value of Standard Angle(30,60,90)

If θ is an ac...

Question

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) $\frac{7}{8}$
(b) $\frac{8}{7}$
(c) $\frac{7}{4}$
(d) $\frac{64}{49}$

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

sin θ (1 + tan θ) + c o s θ (1 + cot θ) = csc θ + sec θ

sin θ (1 + tan θ) + cos θ (1 + cot θ) = sec θ + c o s e c θ

cot θ = \frac{7}{8}

\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}

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