Question

If cot=7/8,then find the values of (1+sin)(1-sin)/(1+cos)(1-cos)

Answer 1

(1 +sin)(1-sin) = 1 - sin^2 = cos^2

( 1 + cos)( 1 -cos)= 1 - cos^2 = sin^2

So ( 1 + sin)( 1- sin)/ (1+cos)(1-cos)

= cos^2/sin^2

= cot^2

As cot = 7/8

so cot^2 = 49/64



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Answer 2

Given,
$cot \: \theta = \frac{7}{8}$

We know that,

$\boxed{\mathbf{Cot = \frac{Base}{Perpendicular}}}$

Hence,

$\bf \frac{Base}{Perpendicular} = \frac{7}{8}$

i.e. Base = 7 and Perpendicular = 8

To calculate the value of Hypotenuse,

$\bf {Hypotenuse}^{2} = {Perpendicular}^{2} + {Base}^{2}$

${h}^{2} = {(8)}^{2} + {(7)}^{2} \\ \\ {h}^{2} = 64 + 49 \\ \\ {h}^{2} = 113 \\ \\ \bf h = \sqrt{113}$

Now,

$\frac{(1 + sin \: \theta)(1 - sin \: \theta)}{(1 + cos \: \theta)(1 - cos \: \theta)}$
We know that,

$\boxed{\mathbf{Sin = \frac{Perpendicular}{Hypotenuse}}}$

$\boxed{\mathbf{Cos = \frac{Base}{Hypotenuse}}}$

Putting the values we get,

$\frac{(1 + \frac{8}{ \sqrt{113} })(1 - \frac{8}{ \sqrt{113} }) }{(1 + \frac{7}{ \sqrt{113} })(1 - \frac{7}{ \sqrt{113} } ) }$

Using the identity :

$\boxed{\mathbf{(a + b)(a - b) = {a}^{2} - {b}^{2}}}$

$= \frac{ {(1)}^{2} - {( \frac{8}{ \sqrt{113} } )}^{2} }{ {(1)}^{2} - {( \frac{7}{ \sqrt{113} } )}^{2} } \\ \\ = \frac{1 - \frac{64}{113} }{1 - \frac{49}{113} } \\ \\ = \frac{1 \times 113 - 64 \times 1}{113} \div \frac{1 \times 113 - 49 \times 1}{113} \\ \\ = \frac{113 - 64}{113} \div \frac{113 - 49}{113} \\ \\ = \frac{49}{113} \div \frac{64}{113} \\ \\ = \frac{49}{113} \times \frac{113}{64}$

$\boxed{\boxed{\mathbf{= \frac{49}{64}}}}$