Question

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Answer 1

Given:

$\tan \theta \: = \frac{8}{15}$

To show that,

$\sqrt{ \frac{1 - sin \theta}{1 + sin \theta} } = \frac{3}{5}$

Solution:

We know that

$\boxed{ \tan\theta = \frac{Perpendicular}{Base} }$

We are given that $tan \theta = \frac{8}{15}$

On comparing

Perpendicular (P) = 8

Base (B) = 15

We can find the Hypotenuse(H) by using Pythagoras Theorem

$\boxed{H {}^{2} = {P}^{2} + {B}^{2} }$

${H}^{2} = (8) {}^{2} + {(15)}^{2} \\ h {}^{2} = 64 + 225 \\ {H}^{2} = 289 \\ H = \sqrt{289} \\ H = 17 \\ Hypotenuse (H) = 17$

$\boxed {sin \theta = \frac{Perpendicular}{Hypotenuse} }$

So, $sin \theta = \frac{8}{17}$

We can solve it by two ways,

• First by putting the values

$\sqrt{ \frac{1 - sin \theta}{1 + sin \theta} }$

$= > \large {\sqrt{ \frac{1 - \frac{8}{17} }{1 + \frac{8}{17} } }} \\ \\ = > \large{\sqrt{ \frac{ \frac{17 - 8}{17} }{ \frac{17 + 8}{17} } }} \\ \\ = > \large{\sqrt{ \frac{ \frac{9}{17} }{ \frac{25}{17} }} } \\ \\ = > \large{ \sqrt{ \frac{ \frac{9}{ \cancel{17}} }{ \frac{25}{ \cancel{17}} } } } \\ \\ = > \sqrt{ \frac{9}{25} } \green{= \frac{3}{5} } \\ \\ \texttt{We know that } \sqrt{9} = 3 \\ \texttt{and} \sqrt{25} = 5 \\ \\ \blue{\implies \sqrt{ \frac{1 - sin \theta}{1 + sin \theta} } = \frac{3}{5} }$

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• Second by simplifying

$\sqrt{ \frac{1 - sin \theta}{1 + sin \theta} }$

Multiplying numerator and denominator by $(1 - \sin \theta)$

$\implies \: \sqrt{ \frac{1 - \sin \theta \times (1 - \sin \theta) }{1 + \sin \theta \times (1 - \sin \theta) } } \\ \\ = > \sqrt{ \frac{(1 - \sin \theta) {}^{2} }{( {1}) ^{2} - ( \sin \theta) {}^{2} } } \\ \\ = > \sqrt{ \frac{(1 - \sin \theta) {}^{2} }{1 - { \sin }^{2} \theta } }$

We know that

(a+b)(a-b) = a²-b²

Using identity

$\sin {}^{2} \theta+ \cos {}^{2} \theta = 1 \\ \therefore \\ \cos {}^{2} \theta = 1 - { \sin }^{2} \theta$

$\implies \: \sqrt{ \frac{(1 - \sin \theta) {}^{2} } {\cos {}^{2} \theta} } \: \\ \\ \implies \: \sqrt{ {\left(\frac{1 - \sin \theta} {cos \theta} \right)} ^{2} } \\ \\ \implies \frac{1 - \sin \theta }{cos \theta} \\ \\ = = > \frac{1}{ \cos \theta} - \frac{ \sin \theta }{ \cos \theta} \\ \\ \left(\because \: \frac{1}{ \cos \theta } = \sec \theta \right) \texttt{and} \\ \\ \left( \frac{ \sin \theta }{ \cos \theta } = \tan \theta \right)\\ \\ \large = \underline{ \sec \theta - \tan \theta }\: \: .....(1)$

Now we have

$\red{\tan \theta = \frac{8}{15} }$

and we know that

$\boxed{ \sec \theta = \frac{Hypotenuse}{Base}} \\ \\ \red{ \implies \sec \theta = \frac{17}{15} }$

Now we will substitute the values in (1)

$= > \frac{17}{15} - \frac{8}{15} \\ \\ = > \frac{9}{15} \: \: \: \: \green{ = \frac{3}{5} } \\ \\ \\ \blue{\implies { \sqrt{ \frac{1 - \sin \theta}{1 + \sin \theta} } } = \frac{3}{5} }$

Answer 2

Answer:

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