Question

If cot θ = 7/8, the evaluate {(1 + sin θ)*(1 - sin θ)}/{(1 + cos θ)*(1 - cos θ)}​

Answer 1

let theta =x.

given,

cot x=7/8

=adjacent side/opposite

side

hypotenuse=

$\sqrt{(7)^{2} } + (8) {}^{2}$

=

$\sqrt{49 + 64}$

=

$\sqrt{113}$

Sin x=8/

$\sqrt{113}$

sin^2 x= 64/113

cos x=7/

$\sqrt{113}$

cos ^2 x =49/113

=(1+sin x)(1-sin x)/(1+cos x)(1- cosx)

=(1-sin^2 x)/(1-cos^2 x)

=(1-64/113)/(1-49/113)

=(113-64/113)/(113-49/113)

=(49/113)/(64/113)

=49/64

(or)

=(7/8)^2

Answer 2

Given:

• $\sf { cot \theta = \dfrac{7}{8} }$

We have to evaluate,

• $\sf { \dfrac{(1 + \sin\theta)(1 - \sin \theta)}{(1 +\cos\theta)(1 - \cos \theta) } }$

Calculation:

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

$\sf { : \implies \dfrac{1 - \cancel{\sin \theta} + \cancel{\sin \theta} - {\sin}^{2} \theta }{1 - \cancel {\cos \theta} +\cancel{ \cos \theta} - { \cos}^{2} \theta} }$

$\sf { : \implies \dfrac{1 - { \sin }^{2} \theta}{1 - { \cos}^{2} \theta} }$

NOW, let ABC be a triangle , right angled at B

Then,

$\sf { : \implies cot \theta = \dfrac{7}{8} ( Given )}$

$\sf { : \implies \dfrac{Base}{Perpendicular} =\dfrac{BC}{AB}= \dfrac{7}{8} }$

Let AB be 8x and BC be 7x

Applying pythagoras property-

$\sf\red { {(H)}^{2} = {(B)}^{2} + {(P)}^{2} }$

$\sf { \longrightarrow {(H)}^{2} = {(7x)}^{2} + {(8x)}^{2} }$

$\sf { \longrightarrow {(H)}^{2} = {49x}^{2} + {64x}^{2} }$

$\sf { \longrightarrow {(H)}^{2} = {113x}^{2} }$

$\sf { \longrightarrow H = \sqrt{{113x}^{2}} = x \sqrt{113} }$

Perpendicular = 8x

Base = 7x

Hypotenuse = $\sf{ x \sqrt{113}}$

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Now we have to find the value of

• $\sf {\dfrac{1 - { \sin }^{2} \theta}{1 - { \cos}^{2} \theta} ( Equation \: 1 ) }$

$\sf {:\implies \sin \theta = \dfrac{P}{H} = \dfrac{ 8 \cancel x}{ \cancel x \sqrt{113} } }$

$\sf {:\implies \cos \theta = \dfrac{B}{H} = \dfrac{ 7 \cancel x}{ \cancel x \sqrt{113 } } }$

Now put the value of $\rm { \sin \theta }$ and $\rm { \cos \theta }$ in the 1st equation.

$\sf\blue { : \implies \dfrac{1 - ( { \dfrac{8}{ \sqrt{113} } )}^{2} }{1 - ( { \dfrac{7}{ \sqrt{113} } )}^{2}} }$

$\sf { : \implies \dfrac{1 - \dfrac{64}{113} }{1 - \dfrac{49}{113} } }$

$\sf { : \implies \dfrac{ \dfrac{113 - 64}{113} }{ \dfrac{113 - 49}{113} } }$

$\sf { : \implies \dfrac{ \dfrac{49}{\cancel {113}} }{ \dfrac{64}{\cancel {113}} } }$

$\sf\red { : \implies \dfrac{49}{64} \: Ans.}$

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