Answer:
L.S = R.S ⇒ Proved down
Step-by-step explanation:
Let us revise some rules in trigonometry
- sin²α + cos²α = 1
- sin2α = 2 sin α cosα
- cscα = 1/sinα
To solve the question let us find the simplest form of the right side and the left side, then show that they are equal
∵ L.S = csc2α + 1
→ By using the 3rd rule above
∴ L.S = [tex]\frac{1}{sin2\alpha}[/tex] + 1
→ Change 1 to [tex]\frac{sin2\alpha}{sin2\alpha}[/tex]
∴ L.S = [tex]\frac{1}{sin2\alpha}[/tex] + [tex]\frac{sin2\alpha}{sin2\alpha}[/tex]
→ The denominators are equal, then add the numerators
∴ L.S = [tex]\frac{1+sin2\alpha}{sin2\alpha}[/tex]
∵ R. S = [tex]\frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}[/tex]
∵ (sinα + cosα)² = sin²α + 2 sinα cosα + cos²α
∴ (sinα + cosα)² = sin²α + cos²α + 2 sinα cosα
→ By using the 1st rule above, equate sin²α + cos²α by 1
∴ (sinα + cosα)² = 1 + 2 sinα cosα
→ By using the 2nd rule above, equate 2 sinα cosα by sin2α
∴ (sinα + cosα)² = 1 + sin2α
→ Substitute it in the R.S above
∴ R. S = [tex]\frac{1+sin2\alpha}{sin2\alpha}[/tex]
∵ L.S = R.S
∴ csc 2α + 1 = [tex]\frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}[/tex]
