Question

sin^3A+cos^3A/sinA+cosA=1-sin*cos prove it

Answer 1

Answer:

Step-by-step explanation:

(sin ^3 A + cos^3 A)/(sinA + cosA)  

or (sin ^2 A *sinA + cos^2 A*cosA)/(sinA + cosA)  

or ((1-cos^2 A)*sinA + (1-sin^2 A)*cosA)/(sinA + cosA)  

or (sinA-cos^2 AsinA + cosA-sin^2 AcosA)/(sinA + cosA)  

or (sinA + cosA - cos^2 AsinA - sin^2 AcosA)/(sinA + cosA)  

or (sinA + cosA - cos^2 AsinA - cosAsin^2 A)/(sinA + cosA)  

or ((sinA + cosA) - cosAsinA*(cosA + sinA))/(sinA + cosA)  

or ((sinA + cosA) - cosAsinA*(sinA + cosA))/(sinA + cosA)  

or ((sinA + cosA)*(1 - cosAsinA))/(sinA + cosA)  

or (sinA + cosA)*(1 - cosAsinA)/(sinA + cosA)  

or (1 - cosAsinA)

Answer 2

Step-by-step explanation:

Answer:

Step-by-step explanation:

Diameter of the metallic sphere = 9 cm

Radius of the metallic sphere , r = 9/2 cm = 4.5 cm

Volume of the metallic sphere = 4/3 × πr³

Diameter of the cylindrical wire = 2 mm = 2/10 cm = 0.2 cm

Radius of the cylindrical wire , r1 = 0.2/2 cm = 0.1 cm

Let the height of the cylindrical wire = h cm

Volume of the cylindrical wire = πr1²×h

Volume of the metallic sphere = Volume of the cylindrical wire

4/3 × πr³ = πr1²×h

4/3 × π  × 4.5³ = π(0.1)²×h

4/3 × 4.5 × 4.5 × 4.5 = 0.01h

4 × 1.5 ×  4.5 × 4.5 = 0.01 × h

h = (4 × 1.5 ×  4.5 × 4.5)/0.01

h = 121.5/0.01 = 121.5 × 100 = 12150 cm

h = 12150 cm  

Hence, the required length of the wire is 12150 cm.

Answer:

Step-by-step explanation:

(sin ^3 A + cos^3 A)/(sinA + cosA)  

or (sin ^2 A *sinA + cos^2 A*cosA)/(sinA + cosA)  

or ((1-cos^2 A)*sinA + (1-sin^2 A)*cosA)/(sinA + cosA)  

or (sinA-cos^2 AsinA + cosA-sin^2 AcosA)/(sinA + cosA)  

or (sinA + cosA - cos^2 AsinA - sin^2 AcosA)/(sinA + cosA)  

or (sinA + cosA - cos^2 AsinA - cosAsin^2 A)/(sinA + cosA)  

or ((sinA + cosA) - cosAsinA*(cosA + sinA))/(sinA + cosA)  

or ((sinA + cosA) - cosAsinA*(sinA + cosA))/(sinA + cosA)  

or ((sinA + cosA)*(1 - cosAsinA))/(sinA + cosA)  

or (sinA + cosA)*(1 - cosAsinA)/(sinA + cosA)  

or (1 - cosAsinA)

hope this helps you!!