Question

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❤ Prove the indentity , Where the angles involved are acute angles for which of the following expressions are defined :-

$\frac{CosA - SinA + 1 }{CosA + SinA -1 } = Cosec A + CotA$


Using the identity
$Cosec^{2} A = 1 + Cot^{2} A$
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⬛@Brainlestuser , Brainly Team , Moderator

Answer 1

$Given : \frac{cosA - sinA + 1}{cosA + sinA - 1}$

Divide both the numerator and denominator by sinA, we get

$= > \frac{\frac{cosA}{sinA} - \frac{sinA}{sinA} + \frac{1}{sinA}}{\frac{cosA}{sinA} + \frac{sinA}{sinA} - \frac{1}{sinA}}$

$= > \frac{cotA - 1 + cosecA}{cotA + 1 - cosecA}$

$= > \frac{cotA + cosecA - 1}{cotA - cosecA + 1}$

Given cosec^2A = 1 + cot^2A. It can be written as cosec^2A - cot^2A = 1

$\frac{cotA + cosecA - (cosec^2A - cot^2A)}{cotA - cosecA + 1}$

We know that a^2 - b^2 = (a + b)(a - b)

$= > \frac{cotA + cosecA - (cosecA + cotA)(cosecA - cotA)}{(cotA - cosecA + 1)}$

$= > \frac{(cosecA + cotA)[1 - cosecA + cotA]}{(1 - cosecA + cotA)}$

$= > cosecA + cotA$

Note: It can be solved using conjugate method also. But we cannot use identity.

Hope it helps!

Answer 2

$<b> <I>$

Hey there !!


$\boxed{ See \: the \: attachment \: for \: your \: answer. }$



$\large \boxed{ \mathbb{TRIGONOMETRY.}}$


$\huge \bf \underline{ \mathbb{LHS = RHS.}}$


✔✔ Hence, it is proved ✅✅.

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$\huge \boxed{ \mathbb{THANKS}}$



$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$