Question

if sinA-cosA=1/2 .then find 1/sinA+cosA

Answer 1

Answer:

1/2 is wrong answer.

1/(sinA + cosA) = 2/√7

Step-by-step explanation:

Given that sin(A) - cos(A) = 1/2

Square both sides, you get

(sin(A) - cos(A))^2 = 1/4

sin^2(A) + cos^(A) - 2sin(A)cos(A) = 1/4

Sin^2(A) + cos^2(A) = 1.

Substituting in above equation

2Sin(A)Cos(A) = 1 - 1/4 = 3/4

Now Take the equation

(Sin(A) + Cos(A))^2 = sin^2(A) + cos^(A) + 2sin(A)cos(A)

= 1 + 2Sin(A)Cos(A) = 1 + 3/4 = 7/4

Apply square root both sides, you get

Sin(A) + Cos(A) = √7/2

Hence 1/(Sin(A) + Cos(A)) = 2/√7

Answer 2

Answer:

$\frac{1}{SinA+CosA}=\frac{2}{\sqrt{7} }$

Step-by-step explanation:

First we need to know the following TRIGONOMETRIC IDENTITY:

$sin^{2}A+cos^{2}A=1$

Now we need to find an equality on our first equation.

If, $sinA-cosA=\frac{1}{2}$, we can square both sides of the equation resulting in:

$sin^{2}A-2sinAcosA+cos^{2}A =\frac{1}{4}$

$sin^{2}A+cos^{2}A-2sinAcosA=\frac{1}{4}$

Replacing the TRIGONOMETRIC IDENTITY:

$1-2sinAcosA=\frac{1}{4}$

$-2sinAcosA=\frac{1}{4}-1$

$2sinAcosA=1-\frac{1}{4}=\frac{3}{4}$  

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Now we need to transform the second equation, REMEBER we can square a number and root-extract it at the same time without altering its value.

$\sqrt{\frac{1 }{(sinA+cosA)^{2} } } =\sqrt{\frac{1 }{sin^{2}A+2senAcosA+cos^{2}A} } =\sqrt{\frac{1 }{sin^{2}A+cos^{2}A+2senAcosB} }$

Now we can use the equality we found previously and the TRIGONOMETRIC IDENTITY to find our answer:

$\sqrt{\frac{1 }{sin^{2}A+cos^{2}A+2senAcosB} }= \sqrt{\frac{1 }{(1)+(\frac{3}{4})}}=\sqrt{\frac{1 }{(\frac{7}{4}) } }=\sqrt{\frac{4 }{7} }= \frac{2}{\sqrt{7} }$