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LHS = 1 - [ (sin²x tanx ) / (tanx + 1) ] - [ cos²x / (tanx + 1) ] = [( tanx...

2 Answers · Science & Mathematics · 08/12/2017

2. ### verify the identity cot(x-pi/2)=-tan x?

LHS= cos(x-pi/2)/sin(x-pi/2)= [cos(x)*0+sin(x)*1]/ [sin(x)*0-cos(x)*1]= sin(x)/[-cos(x)]= -tan(x)= RHS.

4 Answers · Science & Mathematics · 12/01/2018

3. ### can someone help me with my math homework?

...(θ/2)) = |cos(θ/2)| = cos(θ/2) ..... since cos(θ/2) > 0 for 0 ≤ θ ≤ π = LHS

3 Answers · Science & Mathematics · 03/12/2017

4. ### Help with trig identities?

.../cosA 1 - sin²A = cos²A Starting with the LHS we have : ∴ LHS = [(secA + tanA) /(secA - tanA...

4 Answers · Science & Mathematics · 20/11/2017

5. ### Prove this trig identity?

LHS = (secA-cosA)/(cosA-sinA) = (1 - cos²A...

2 Answers · Science & Mathematics · 19/11/2017

6. ### Establish the following identity: csc(theta) - cot(theta) = (sin(theta))/(1+cos(theta))?

LHS: csc θ - cot θ = 1/sin θ - cos θ/sin θ = (1 - cos θ)/sin θ Multiply...

1 Answers · Science & Mathematics · 08/11/2017

7. ### Prove math 10 points?

... - cos2A = 2sin(B + A)sin(A - B) LHS = [2cos(A + B)cos(A - B)] /[2sin(B + A)sin...

1 Answers · Science & Mathematics · 21/10/2017

8. ### How to proof this identity trigonometry?

LHS= cos(4x)+cos(6x)-cos(2x)-1= cos(4x)+cos(6x)-2cos^2(x)= 2cos(5x)cos(x)-2cos^2(x)= 2cos(x)[cos(5x)-cos(x)]= 2cos(x)[-2sin(3x)sin(2x)]= -4cos(x)sin(3x)sin(2x)= RHS

2 Answers · Science & Mathematics · 21/10/2017

9. ### How to solve 5/6x=-2/18 What are the steps? Please explain this example the best you can with steps to your final answer.?

5/6x=-2/18 multiply LHS by 3/3 15/18x = -2/18 multiply both sides by 18 15x = -2 divide both ...

6 Answers · Science & Mathematics · 16/10/2017

10. ### Prove this identity trigonometry?

... + sin B) = (sin B - sin A) / (cos A + cos B) LHS = (cosA - cosB)/(sinA + sinB) {multiply by conjugate of...

4 Answers · Science & Mathematics · 14/10/2017