Question

pre baord question

class 10​

Answer 1

Answer:

L.H.S= √1-cosA/1+cosA

= √1-cosA. 1-cosA

----------------------- ×-------------------

1+cosA. 1-cosA

√(1+cosA)²

------------------

1-cosA²

√(1+cosA)²

-----------------

sin²A

=. 1-cosA

-----------

sinA

cosecA- cotA= R.H.S

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Answer 2

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Prove \: that \: : \sqrt{\dfrac{1 + cosA}{1 - cosA} } = cosecA - cotA$

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

Identities used :-

$(1). \: \boxed{ \bf{1 - {cos}^{2} A = {sin}^{2}A }}$

$(2). \: \boxed{ \bf{cosecA = \dfrac{1}{sinA} }}$

$(3). \: \boxed{ \bf{\dfrac{cosA}{sinA} = cotA}}$

Let's solve the problem now!!

Consider LHS

$\sf \: \sqrt{\dfrac{1 + cosA}{1 - cosA} }$

On rationalizing the denominator, we get

$\sf \: = \sqrt{\dfrac{1 + cosA}{1 - cosA} \times \dfrac{1 + cosA}{1 + cosA} }$

$\sf \: = \sqrt{\dfrac{ {(1 + cosA)}^{2} }{1 - {cos}^{2}A } }$

$\sf \: = \sqrt{\dfrac{ {(1 + cosA)}^{2} }{ {sin}^{2} A} }$

$\sf \: = \dfrac{1 + cosA}{sinA}$

$\sf \: = cosecA + cotA$

$\bf \: = RHS$

$\large{{\boxed{\bf{Hence, Proved}}}}$

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