Question

If cos A-sin A=1 prove that cos A+sin A =1

Answer 1

♧♧HERE IS YOUR ANSWER♧♧



Trigonometry is the study of angles and its ratios. This study prescribes the relation amongst the sides of a triangle and angles of the triangle.



There are sine, cosine, tangent, cot, sectant and cosec ratios to an angle of a triangle.



Let me tell you an interesting fact about Trigonometry.



"Triangle" > "Trigonometry"



Remember some formulae now :



sin²θ + cos²θ = 1



sec²θ - tan²θ = 1



cosec²θ - cot²θ = 1



Want to learn more!



Here it is :



sin(A + B) = sinA cosB + cosA sinB



sin(A - B) = sinA cosB - cosA sinB



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



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




SOLUTION :

Given :

cosA - sinA = 1

Squaring both sides we get :

(cosA - sinA)² = 1²

=> cos²A - 2cosA sinA + sin²A = 1

=> (cos²A + sin²A) - 2cosA sinA = 1

=> 1 - 2cosA sinA = 1

Now, cancelling 1, we get :

2cosA sinA = 0 .....(i)

Now,

(cosA + sinA)²

= cos²A + 2cosA sinA + sin²A

= cos²A + sin²A + 2×0, by (i)

= 1

Since, sin²A + cos²A always values 1,

cosA + sinA = 1, ≠ -1.

Therefore?

cosA + sinA = 1

Hence, proved.

♧♧HOPE THIS HELPS YOU♧♧

Answer 2

Hey friend, Harish here.

Here is your answer:


Given that,

$cosA - sinA= 1 \ \ \ \ \ \  - (i)$

To prove,

$cos A+sin A =1$

Proof,

We know that,

$(a-b)^{2} + (a+b)^{2} = 2(a^{2} + b^{2})$

$sinA^{2} + cosA^{2} = 1 \ \ \ \ \ - (ii)$   

Here, 

$a = cosA\ , \ b =sinA$

Then,

$(cosA-sinA)^{2} + (cosA+sinA)^{2} = 2( sinA^{2} + cosA^{2})$

Now, Substitute value of (i) and (ii) in the above equation.

Then

→ $1^{2} +(cosA+sinA)^{2} = 2 \times 1$

→ $(cosA+sinA)^{2} = 2-1$

→ $(cosA+sinA) = \sqrt{1}$

→ $(cosA+sinA) = 1$

Hence proved that cosA + sinA = 1.
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Hope my answer is helpful to you.