Question

if tan ø=7/8,then what is (1+sinø)(1-sinø)/(1+cosø)(1-cosø)​

Answer 1

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▪ Given :-

$\bf{ \tan \phi = \dfrac{7}{8} }$

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▪To Find :-

$\dfrac{(1 + \sin \phi)(1 - \sin\phi)}{(1 + \cos\phi)(1 - \cos\phi)}$

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▪ Used Formulae :-

$\pmb{ \maltese \: \: \: (x + y)(x - y) = {x}^{2} - {y}^{2} } \\ \\ \pmb{ \maltese \: \: \: \sin {}^{2} \theta + { \cos}^{2} \theta = 1} \\ \\ \pmb{ \maltese \: \: \: \dfrac{ \cos \theta}{ \sin \theta } = \cot \theta} \\ \\ \pmb{ \maltese \: \: \: \cot \theta = \dfrac{1}{ \tan \theta} }$

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▪ Solution :-

We Have,

$\dfrac{(1 + \sin \phi)(1 - \sin\phi)}{(1 + \cos\phi)(1 - \cos\phi)} \\ \\ =\frac{1 {}^{2} - { \sin}^{2} \phi}{1 {}^{2} - { \cos}^{2} \phi } \\ \\ = \frac{1 - { \sin}^{2} \phi}{1 - { \cos}^{2} \phi } \\ \\ = \frac{ { \cos}^{2} \phi}{ { \sin}^{2} \phi } \\ \\ = { \cot}^{2} \phi \\ \\ = \frac{ 1}{ { \tan}^{2} \phi } \\ \\ = \frac{ \: \: \: 1 \: \: \: }{ \big( \frac{7}{8} \big) {}^{2} } \\ \\ = \bigg(\frac{8 { }^{} }{7} \bigg) {}^{ {}^{ {}^{ {}^{ {}^{2} } } } } \\ \\ = \pmb{ \dfrac{64}{49}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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$\large\bf Fundamental \: Trigonometric \\ \Large \bf Identities \\ \\ \maltese \: \: \:\sin^2\theta + \cos^2\theta=1 \\ \\ \maltese \: \: \: 1+\tan^2\theta = \sec^2\theta \\ \\\maltese \: \: \: 1+\cot^2\theta = \text{cosec}^2 \, \theta$

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Answer 2

$\dfrac{(1 + \sin \phi)(1 - \sin\phi)}{(1 + \cos\phi)(1 - \cos\phi)} \\ \\ =\frac{1 {}^{2} - { \sin}^{2} \phi}{1 {}^{2} - { \cos}^{2} \phi } \\ \\ = \frac{1 - { \sin}^{2} \phi}{1 - { \cos}^{2} \phi } \\ \\ = \frac{ { \cos}^{2} \phi}{ { \sin}^{2} \phi } \\ \\ = { \cot}^{2} \phi \\ \\ = \frac{ 1}{ { \tan}^{2} \phi } \\ \\ \bigg(\frac{8 { }^{} }{7} \bigg) {}^{ {}^{ {}^{ {}^{ {}^{2} } } } } \\ \\ =\mathfrak{\pink{ \LARGE\dfrac{64}{49}}}$