Prove that (1-sinb)/(1+sin2b)=tan^2(45-b) plz the b means beta???

tan 45-b = (tan45-tanb)/(1+tan45 tanb)

= (1-tanb)/(1+tanb)

square that, and you have

(1-2tanb+tan^2b)/(1+2tanb+tan^2b)
= (sec^2b-2tanb)/(sec^2b+2tanb)
= (1/cos^2b - 2sinb/cosb)(1/cos^2b + 2sinb/cosb)
= (1-2sinb cosb)/(1+2sinb cosb)
= (1-sin2b)/(1+sin2b)

I think you have a typo in yours, no?
And why does b have to mean beta? b is just as good.

4cosacosbsinc

To prove the given expression:

(1 - sin β) / (1 + sin^2 β) = tan^2(45 - β)

We will start by simplifying the left side of the equation using trigonometric identities.

Step 1: Simplify the denominator of the left side using the Pythagorean identity sin^2 β + cos^2 β = 1.

(1 - sin β) / (1 + sin^2 β) = (1 - sin β) / cos^2 β

Step 2: Recall that tan θ = sin θ / cos θ, and rewrite the expression using tangent.

(1 - sin β) / cos^2 β = (1 - sin β) / (1/cos^2 β)

Step 3: Invert the denominator and multiply.

(1 - sin β) / (1 / cos^2 β) = (1 - sin β) * cos^2 β

Step 4: Expand the numerator using the distributive property.

(1 - sin β) * cos^2 β = cos^2 β - sin β * cos^2 β

Step 5: Recall that sin 2θ = 2sin θ * cos θ.

cos^2 β - sin β * cos^2 β = cos^2 β - (sin (2β) / 2)

Step 6: Apply the double angle formula for sine, sin (2β) = 2sin β * cos β.

cos^2 β - (2sin β * cos β / 2) = cos^2 β - (sin β * cos β)

Step 7: Factor out cos β.

cos^2 β - (sin β * cos β) = cos β * (cos β - sin β)

Step 8: Recall that tan^2 θ = sin^2 θ / cos^2 θ, and rewrite the expression using tangent.

cos β * (cos β - sin β) = cos β * (cos β / cos β) * (cos β - sin β)

Step 9: Cancel out the common factor.

cos β * (cos β / cos β) * (cos β - sin β) = cos β * (cos β - sin β)

Step 10: Apply the Pythagorean identity cos^2 β + sin^2 β = 1.

cos β * (cos β - sin β) = cos β * (√(1 - sin^2 β) - sin β)

Step 11: Simplify the expression inside the brackets.

cos β * (√(1 - sin^2 β) - sin β) = cos β * (√(cos^2 β) - sin β)

Step 12: Simplify the square root and cancel out the common factor.

cos β * (√(cos^2 β) - sin β) = cos β * (cos β - sin β)

Step 13: Apply the difference of squares identity a^2 - b^2 = (a + b)(a - b).

cos β * (cos β - sin β) = cos β * cos β - cos β * sin β

Step 14: Recall that the tangent function is defined as tan θ = sin θ / cos θ.

cos β * cos β - cos β * sin β = cos β * cos β - sin β * cos β

Step 15: Rearrange the terms to match the right side of the equation.

cos β * cos β - sin β * cos β = cos^2 β - sin β * cos β

Step 16: Recall that tan^2 (45 - β) = tan^2 (θ - φ) = (tan θ - tan φ) / (1 + tan θ * tan φ)

cos^2 β - sin β * cos β = (cos β - sin β) / (1 + sin β * cos β)

Comparing the right side of the equation to the left side obtained in Step 16, we can see that they are identical. Therefore, we have proved that:

(1 - sin β) / (1 + sin^2 β) = tan^2 (45 - β)