Question

cos (45º -A) - cos (45° + A
(A) √2 cos A
(C) √2 sin A
(B) 12 sin A
(D) 2 cos A​

Answer 1

$\sf{Answer}$

Step by step explanation:-

Given to find value of :-

cos (45º -A) - cos (45° + A)

Formulae to know :-

• cos (A - B ) = cosA cosB + sinAsinB
• cos ( A + B ) = cosA cosB - sinA sinB

sin45° = 1/√2

$\huge\bf{Solution}$

• cos (45- A ) In the form of cos (A - B)
• cos (45 + A ) in the form of cos (A + B )

Lets solve !

cos (45º -A) - cos (45° + A)

cos45 cosA + sin45 sinA - [ cos45 cosA - sin45 sinA]

cos45 cosA + sin45 sinA - cos45cosA + sin45 sinA

Keep like terms together

cos45cosA - cos45 cosA + sin45 sinA + sin45 sinA

sin45 sinA + sin45 sinA

1/√2 sinA + 1/√2 sinA

2/√2 sinA

√2 sinA

So, cos (45º -A) - cos (45° + A) = √2 sinA

Know more :-

tan ( A +B ) = tanA + tanB / 1 - tanAtanB

tan( A-B) = tanA - tanB/1 + tanAtanB

cot ( A + B ) = cotBcotA -1 / cotB + cotA

cot ( A - B ) = cotB cotA + 1/ cotB - cotA

tan(45+ A) = 1+tanA/1 - tanA

tan (45 - A ) = 1-tanA/1 + tanA

Trignometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trignometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trignometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj

Answer 2

$\huge \sf \underline{Answer}:$

$\large \tt \implies (C) \sqrt{2}sinA$

$\huge \sf \underline{Solution}:$

$\large \sf Solve :$
$\: \: \: \: \: \: \: \: \: \:$➽ cos (45º -A) - cos (45° + A)

We know ,

$\sf \large \boxed{\sf cos x - cos y = -2sin \frac{x+y}{2} sin \frac{x-y}{2} }$

$\large \sf Take :$
$\: \: \: \: \: \: \: \: \: \:$ ● Cos(45° - A) = cos x
$\: \: \: \: \: \: \: \: \: \:$ ● Cos(45° + A) = cos y

Now, Apply the formula here,

cos (45º -A) - cos (45° + A) =

$\: \: \: \: \: \:$ $\sf -2sin (\frac{45-\cancel{A}+45+\cancel{A}}{2}) \times sin (\frac{45-A-(45+A)}{2} )$
$\\ \implies \sf -2sin( \frac{90}{2} )\times sin (\frac{\cancel{45}-A-\cancel{45}-A}{2})$
$\\ \sf \implies -2sin 45 (-sin (\frac{2A}{2})) \\ \\ \implies \sf 2sin 45 sin A$

$\large \boxed{Sin45 = \frac{1}{\sqrt{2}}}$

$\implies \sf 2\times \frac{1}{\sqrt{2}} sin A \\ \sf \implies \frac{2}{\sqrt{2}} sin A$

$\large \bf Rationalising$

$\implies \sf \frac{2}{\sqrt{2}} sin A \times \frac{\sqrt{2}}{\sqrt{2}} \\ \\ \sf \implies \sqrt{2}sin A$
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Therefore ,
☞ cos (45º -A) - cos (45° + A) = √2sin A

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