Question

If Cot o= 7/8
evalute (1+sin o) (1-sin o)
(1+ Cos o ) (1-Cos o)​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:cot \theta \: = \: \dfrac{7}{8}$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:\dfrac{(1 + sin\theta)(1 - sin\theta)}{(1 + cos\theta)(1 - cos\theta)}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\green{\boxed{ \bf \: {sin}^{2}x + {cos}^{2}x = 1}}$

$\green{\boxed{ \bf \:\dfrac{cosx}{sinx} = cotx }}$

$\green{\boxed{ \bf \: (x + y)(x - y) = {x}^{2} - {y}^{2} }}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:cot \theta \: = \: \dfrac{7}{8}$

Consider,

$\rm :\longmapsto\:\dfrac{(1 + sin\theta)(1 - sin\theta)}{(1 + cos\theta)(1 - cos\theta)}$

$\rm \: = \: \: \dfrac{1 - {sin}^{2}\theta }{1 - {cos}^{2} \theta}$

$\rm \: = \: \: \dfrac{ {cos}^{2} \theta}{ {sin}^{2} \theta}$

$\rm \: = \: \: {cot}^{2} \theta$

$\rm \: = \: \: ({cot}\theta)^{2}$

$\rm \: = \: \: {\bigg(\dfrac{7}{8} \bigg) }^{2}$

$\rm \: = \: \: \dfrac{49}{64}$

$\bf\implies \:\:\dfrac{(1 + sin\theta)(1 - sin\theta)}{(1 + cos\theta)(1 - cos\theta)} = \dfrac{49}{64}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

Answer 2

this is your answer....