How do I prove that (cosecA-cotA) ²=1-cosA/1+cosA?
Here is the answer.Hope you get it :)
Prove : ((1+cosA)/(1-cosA))^1\2 = cosecA +cotA?
[ (1 + cos(A)) / (1 - cos(A)) ]^(1/2) = csc(A) + cotA LHS = [ (1 + cos(A)) / (1 - cos(A)) ]^(1/2) Mutiply top and bottom by conjugate of bottom. LHS = [ (1 + cos(A))^2 / (1 - cos^2(A)) ]^(1/2) By our trig identities, 1 - cos^2(A) = sin^2(A). LHS = [ (1 + cosA)^2 / [sin^2(A)] ]^(1/2) We have a square value in the numerator and denominator, taking it to the power of (1/2). This means we can effectively remove the squares. LHS = (1 + cos(A)) / sin(A) Which we can split into two fractions, LHS = 1/sin(A) + cos(A)/sin(A) LHS = csc(A) + cot(A) = RHS
How can we prove (cosA-square.root (1+sin2A)) / (sinA-square.root)) =tanA?
The question is incomplete, I'll assume the missing part, we want to prove[math]\frac{cosA - \sqrt{1 + sin2A}}{sinA - \sqrt{1 + sin2A}} = tanA \tag 1[/math]You can write[math] sin2A = 2sinAcosA \tag*{} [/math][math] 1 = sin^2A + cos^2A \tag*{} [/math]Therefore,[math] \sqrt{1 + sin2A} = \sqrt{sin^2A + cos^2A + 2sinAcosA}[/math][math] \implies \sqrt{(sinA + cosA)^2} \tag*{} [/math][math] \implies sinA + cosa \tag*{} [/math]Putting this in LHS of [math](1)[/math][math]\frac{cosA - (sinA + cosA)}{sinA - (sinA + cosA)} [/math][math]\implies \frac{-sinA}{-cosA} = tanA \tag*{} [/math]
How do I prove tha √1+cosA/√1-cosA=cosecA+cotA?
The proof is as follows-To prove- √(1+cosA)/√(1-cosA)=cosecA + cot AProof-LHS-√(1+cosA)/√(1-cosA)Multiplying with √(1+cosA) on both numerator and denominator, we have,={√(1+cosA)*√(1+cosA)}/{√(1-cosA)*√(1+cosA)}=[√{(1+cosA)*(1+cosA)}]/[√{(1-cosA)*(1+cosA)}]={√(1+cosA)²}/{√(1²-cos²A)}=(1+cosA)/√(sin²A)=(1+cosA)/sinA=(1/sinA)+(cosA/sinA)=cosecA + cotAHence proved…
IF tan A/2={(1-e)/(1+e)}1/2 . tan B/2 prove that cos B= (cos A-e)/(1-e cosA)?
I will outline the proof. Square both sides, then multiply both sides by (1+e)/(1-e) (assuming e ≠ 1 -- a case you may want to consider separately). Next, write tan(A/2) = sin(A/2)/cos(A/2) and tan(B/2) = sin(B/2)/cos(B/2). Then use the half-angle identities sin²(x/2) = (1 - cos(x))/2 cos²(x/2) = (1 + cos(x))/2 for both the A/2 and B/2 trig functions. If you've followed that, you should have [(1+e)/(1-e)] [(1 - cos(A))/(1 + cos(A))] = [(1 - cos(B))/(1 + cos(B))] Temporarily let X = [(1+e)/(1-e)] [(1 - cos(A))/(1 + cos(A))], then solve X = [(1 - cos(B))/(1 + cos(B))] for cos(B). You should get cos(B) = (1-X)/(1+X) Now replace X by [(1+e)/(1-e)] [(1 - cos(A))/(1 + cos(A))]: cos(B) = {1 - [(1+e)/(1-e)] [(1 - cos(A))/(1 + cos(A))]} / {1 + [(1+e)/(1-e)] [(1 - cos(A))/(1 + cos(A))]} Multiply the numerator and denominator of the right side by (1-e)(1 + cos(A)) to simplify the fraction. You should get cos(B) = [(1-e)(1 + cos(A)) - (1+e)(1 - cos(A))] / [(1-e)(1 + cos(A)) + (1+e)(1 - cos(A))] Expand the products and combine terms, and you should get the given result.
Prove: sqrt( ( 1-cosA)/(1+cosA) )=sinA/1+cosA?
SQRT [(1 - cos A) /(1 + cos A ) ] multiply top and bottom with SQRT(1 + cos A) = SQRT [ (1 - cos^2(A)) /(1 + cos A)^2 ] = SQRT [ sin A /(1 + cos A) ]^2 = sin A / (1 + cos A)
How do I prove sqrt ((1-cosA)/(1+cosA)) = cosecA -cotA?
Proceeding with LHS. Rationalising LHS and then taking its square root gives sqr root of (1-cosA)^2 /(sinA)^2 which on solving gives RHS
How can I prove this? 1-cos2A/1+cos2A = tan²A
Required To Prove :(1-cos2A) /(1+cos2A) =tan²AProof:We know that,cos(A+B) =cosA.cosB-sinA.sinB=>cos2A=cos(A+A)=>cos2A=cosA.cosA - sinA.sinA=>cos2A=cos²A-sin²A=>cos2A=(cos²A-sin²A)/(cos²A+sin²ASince {cos²A+sin²A=1}Divide the numerator & the denominator by (cos²A) to get,cos2A = {(cos²A-sin²A) ÷cos²A} / {(cos²A+sin²A) ÷cos²A}cos2A ={(1-tan²A)/(1+tan²A)}Then,1-cos2A = 1-[{(1–tan²A)/(1+tan²A)}]1-cos2A =(1+tan²A-1+tan²A)/(1+tan²A)1-cos2A=(2tan²A)/(1+tan²A)And now.......1+cos2A=1+[{(1-tan²A)/(1+tan²A)}]1+cos2A={1+tan²A+1-tan²A}/{1+tan²A}1+cos2A=2/(1+tan²A)So now,(1-cos2A)/(1+cos2A)= {2tan²A/(1+tan²A)}÷{2/(1+tan²A)}={(2tan²A)(1+tan²A)}÷{2(1+tan²A)}=tan²AHENCE PROVED.......